743 lines
21 KiB
Markdown
743 lines
21 KiB
Markdown
<!-- GFM-TOC -->
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* [前言](#前言)
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* [1. 小米-小米Git](#1-小米-小米git)
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* [2. 小米-懂二进制](#2-小米-懂二进制)
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* [3. 小米-中国牛市](#3-小米-中国牛市)
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* [4. 微软-LUCKY STRING](#4-微软-lucky-string)
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* [5. 微软-Numeric Keypad](#5-微软-numeric-keypad)
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* [6. 微软-Spring Outing](#6-微软-spring-outing)
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* [7. 微软-S-expression](#7-微软-s-expression)
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* [8. 华为-最高分是多少](#8-华为-最高分是多少)
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* [9. 华为-简单错误记录](#9-华为-简单错误记录)
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* [10. 华为-扑克牌大小](#10-华为-扑克牌大小)
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* [11. 去哪儿-二分查找](#11-去哪儿-二分查找)
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* [12. 去哪儿-首个重复字符](#12-去哪儿-首个重复字符)
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* [13. 去哪儿-寻找Coder](#13-去哪儿-寻找coder)
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* [14. 美团-最大差值](#14-美团-最大差值)
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* [15. 美团-棋子翻转](#15-美团-棋子翻转)
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* [16. 美团-拜访](#16-美团-拜访)
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* [17. 美团-直方图内最大矩形](#17-美团-直方图内最大矩形)
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* [18. 美团-字符串计数](#18-美团-字符串计数)
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* [19. 美团-平均年龄](#19-美团-平均年龄)
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* [20. 百度-罪犯转移](#20-百度-罪犯转移)
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* [22. 百度-裁减网格纸](#22-百度-裁减网格纸)
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* [23. 百度-钓鱼比赛](#23-百度-钓鱼比赛)
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* [24. 百度-蘑菇阵](#24-百度-蘑菇阵)
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<!-- GFM-TOC -->
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# 前言
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省略的代码：
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```java
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import java.util.*;
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```
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```java
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public class Solution {
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}
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```
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```java
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public class Main {
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public static void main(String[] args) {
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Scanner in = new Scanner(System.in);
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while (in.hasNext()) {
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}
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}
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}
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```
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# 1. 小米-小米Git
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- 重建多叉树
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- 使用 LCA
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```java
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private class TreeNode {
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int id;
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List<TreeNode> childs = new ArrayList<>();
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TreeNode(int id) {
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this.id = id;
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}
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}
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public int getSplitNode(String[] matrix, int indexA, int indexB) {
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int n = matrix.length;
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boolean[][] linked = new boolean[n][n]; // 重建邻接矩阵
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < n; j++) {
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linked[i][j] = matrix[i].charAt(j) == '1';
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}
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}
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TreeNode tree = constructTree(linked, 0);
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TreeNode ancestor = LCA(tree, new TreeNode(indexA), new TreeNode(indexB));
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return ancestor.id;
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}
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private TreeNode constructTree(boolean[][] linked, int root) {
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TreeNode tree = new TreeNode(root);
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for (int i = 0; i < linked[root].length; i++) {
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if (linked[root][i]) {
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linked[i][root] = false; // 因为题目给的邻接矩阵是双向的，在这里需要把它转为单向的
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tree.childs.add(constructTree(links, i));
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}
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}
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return tree;
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}
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private TreeNode LCA(TreeNode root, TreeNode p, TreeNode q) {
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if (root == null || root.id == p.id || root.id == q.id) return root;
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TreeNode ancestor = null;
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int cnt = 0;
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for (int i = 0; i < root.childs.size(); i++) {
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TreeNode tmp = LCA(root.childs.get(i), p, q);
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if (tmp != null) {
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ancestor = tmp;
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cnt++;
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}
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}
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return cnt == 2 ? root : ancestor;
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}
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```
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# 2. 小米-懂二进制
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对两个数进行异或，结果的二进制表示为 1 的那一位就是两个数不同的位。
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```java
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public int countBitDiff(int m, int n) {
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return Integer.bitCount(m ^ n);
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}
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```
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# 3. 小米-中国牛市
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背包问题，可以设一个大小为 2 的背包。
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状态转移方程如下：
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```html
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dp[i, j] = max(dp[i, j-1], prices[j] - prices[jj] + dp[i-1, jj]) { jj in range of [0, j-1] } = max(dp[i, j-1], prices[j] + max(dp[i-1, jj] - prices[jj]))
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```
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```java
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public int calculateMax(int[] prices) {
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int n = prices.length;
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int[][] dp = new int[3][n];
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for (int i = 1; i <= 2; i++) {
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int localMax = dp[i - 1][0] - prices[0];
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for (int j = 1; j < n; j++) {
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dp[i][j] = Math.max(dp[i][j - 1], prices[j] + localMax);
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localMax = Math.max(localMax, dp[i - 1][j] - prices[j]);
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}
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}
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return dp[2][n - 1];
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}
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```
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# 4. 微软-LUCKY STRING
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- 斐波那契数列可以预计算；
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- 从头到尾遍历字符串的过程，每一轮循环都使用一个 Set 来保存从 i 到 j 出现的字符，并且 Set 保证了字符都不同，因此 Set 的大小就是不同字符的个数。
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```java
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Set<Integer> fibSet = new HashSet<>(Arrays.asList(1, 2, 3, 5, 8, 13, 21, 34, 55, 89));
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Scanner in = new Scanner(System.in);
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String str = in.nextLine();
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int n = str.length();
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Set<String> ret = new HashSet<>();
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for (int i = 0; i < n; i++) {
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Set<Character> set = new HashSet<>();
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for (int j = i; j < n; j++) {
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set.add(str.charAt(j));
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int cnt = set.size();
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if (fibSet.contains(cnt)) {
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ret.add(str.substring(i, j + 1));
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}
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}
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}
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String[] arr = ret.toArray(new String[ret.size()]);
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Arrays.sort(arr);
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for (String s : arr) {
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System.out.println(s);
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}
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```
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# 5. 微软-Numeric Keypad
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```java
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private static int[][] canReach = {
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{1, 0, 0, 0, 0, 0, 0, 0, 0, 0}, // 0
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{1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, // 1
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{1, 0, 1, 1, 0, 1, 1, 0, 1, 1}, // 2
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{0, 0, 0, 1, 0, 0, 1, 0, 0, 1}, // 3
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{1, 0, 0, 0, 1, 1, 1, 1, 1, 1}, // 4
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{1, 0, 0, 0, 0, 1, 1, 0, 1, 1}, // 5
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{0, 0, 0, 0, 0, 0, 1, 0, 0, 1}, // 6
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{1, 0, 0, 0, 0, 0, 0, 1, 1, 1}, // 7
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{1, 0, 0, 0, 0, 0, 0, 0, 1, 1}, // 8
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{0, 0, 0, 0, 0, 0, 0, 0, 0, 1} // 9
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};
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private static boolean isLegal(char[] chars, int idx) {
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if (idx >= chars.length || idx < 0) return true;
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int cur = chars[idx] - '0';
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int next = chars[idx + 1] - '0';
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return canReach[cur][next] == 1;
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}
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public static void main(String[] args) {
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Scanner in = new Scanner(System.in);
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int T = Integer.valueOf(in.nextLine());
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for (int i = 0; i < T; i++) {
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String line = in.nextLine();
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char[] chars = line.toCharArray();
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for (int j = 0; j < chars.length - 1; j++) {
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while (!isLegal(chars, j)) {
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if (--chars[j + 1] < '0') {
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chars[j--]--;
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}
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for (int k = j + 2; k < chars.length; k++) {
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chars[k] = '9';
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}
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}
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}
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System.out.println(new String(chars));
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}
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}
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```
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# 6. 微软-Spring Outing
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下面以 N = 3，K = 4 来进行讨论。
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初始时，令第 0 个地方成为待定地点，也就是呆在家里。
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从第 4 个地点开始投票，每个人只需要比较第 4 个地方和第 0 个地方的优先级，里，如果超过半数的人选择了第 4 个地方，那么更新第 4 个地方成为待定地点。
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从后往前不断重复以上步骤，不断更新待定地点，直到所有地方都已经投票。
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上面的讨论中，先令第 0 个地点成为待定地点，是因为这样的话第 4 个地点就只需要和这个地点进行比较，而不用考虑其它情况。如果最开始先令第 1 个地点成为待定地点，那么在对第 2 个地点进行投票时，每个人不仅要考虑第 2 个地点与第 1 个地点的优先级，也要考虑与其后投票地点的优先级。
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```java
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int N = in.nextInt();
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int K = in.nextInt();
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int[][] votes = new int[N][K + 1];
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for (int i = 0; i < N; i++) {
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for (int j = 0; j < K + 1; j++) {
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int place = in.nextInt();
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votes[i][place] = j;
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}
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}
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int ret = 0;
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for (int place = K; place > 0; place--) {
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int cnt = 0;
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for (int i = 0; i < N; i++) {
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if (votes[i][place] < votes[i][ret]) {
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cnt++;
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}
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}
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if (cnt > N / 2) {
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ret = place;
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}
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}
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System.out.println(ret == 0 ? "otaku" : ret);
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```
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# 7. 微软-S-expression
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# 8. 华为-最高分是多少
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```java
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int N = in.nextInt();
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int M = in.nextInt();
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int[] scores = new int[N];
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for (int i = 0; i < N; i++) {
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scores[i] = in.nextInt();
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}
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for (int i = 0; i < M; i++) {
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String str = in.next();
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if (str.equals("U")) {
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int id = in.nextInt() - 1;
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int newScore = in.nextInt();
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scores[id] = newScore;
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} else {
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int idBegin = in.nextInt() - 1;
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int idEnd = in.nextInt() - 1;
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int ret = 0;
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if (idBegin > idEnd) {
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int t = idBegin;
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idBegin = idEnd;
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idEnd = t;
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}
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for (int j = idBegin; j <= idEnd; j++) {
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ret = Math.max(ret, scores[j]);
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}
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System.out.println(ret);
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}
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}
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```
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# 9. 华为-简单错误记录
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```java
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HashMap<String, Integer> map = new LinkedHashMap<>();
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while (in.hasNextLine()) {
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String s = in.nextLine();
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String key = s.substring(s.lastIndexOf('\\') + 1);
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map.put(key, map.containsKey(key) ? map.get(key) + 1 : 1);
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}
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List<Map.Entry<String, Integer>> list = new LinkedList<>(map.entrySet());
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Collections.sort(list, (o1, o2) -> o2.getValue() - o1.getValue());
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for (int i = 0; i < 8 && i < list.size(); i++) {
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String[] token = list.get(i).getKey().split(" ");
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String filename = token[0];
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String line = token[1];
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if (filename.length() > 16) filename = filename.substring(filename.length() - 16);
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System.out.println(filename + " " + line + " " + list.get(i).getValue());
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}
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```
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# 10. 华为-扑克牌大小
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```java
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public class Main {
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private Map<String, Integer> map = new HashMap<>();
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public Main() {
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map.put("3", 0);
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map.put("4", 1);
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map.put("5", 2);
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map.put("6", 3);
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map.put("7", 4);
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map.put("8", 5);
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map.put("9", 6);
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map.put("10", 7);
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map.put("J", 8);
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map.put("Q", 9);
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map.put("K", 10);
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map.put("A", 11);
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map.put("2", 12);
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map.put("joker", 13);
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map.put("JOKER ", 14);
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}
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private String play(String s1, String s2) {
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String[] token1 = s1.split(" ");
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String[] token2 = s2.split(" ");
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CardType type1 = computeCardType(token1);
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CardType type2 = computeCardType(token2);
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if (type1 == CardType.DoubleJoker) return s1;
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if (type2 == CardType.DoubleJoker) return s2;
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if (type1 == CardType.Bomb && type2 != CardType.Bomb) return s1;
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if (type2 == CardType.Bomb && type1 != CardType.Bomb) return s2;
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if (type1 != type2 || token1.length != token2.length) return "ERROR";
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for (int i = 0; i < token1.length; i++) {
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int val1 = map.get(token1[i]);
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int val2 = map.get(token2[i]);
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if (val1 != val2) return val1 > val2 ? s1 : s2;
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}
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return "ERROR";
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}
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private CardType computeCardType(String[] token) {
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boolean hasjoker = false, hasJOKER = false;
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for (int i = 0; i < token.length; i++) {
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if (token[i].equals("joker")) hasjoker = true;
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else if (token[i].equals("JOKER")) hasJOKER = true;
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}
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if (hasjoker && hasJOKER) return CardType.DoubleJoker;
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int maxContinueLen = 1;
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int curContinueLen = 1;
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String curValue = token[0];
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for (int i = 1; i < token.length; i++) {
|
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if (token[i].equals(curValue)) curContinueLen++;
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else {
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curContinueLen = 1;
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curValue = token[i];
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}
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maxContinueLen = Math.max(maxContinueLen, curContinueLen);
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}
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if (maxContinueLen == 4) return CardType.Bomb;
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if (maxContinueLen == 3) return CardType.Triple;
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if (maxContinueLen == 2) return CardType.Double;
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boolean isStraight = true;
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for (int i = 1; i < token.length; i++) {
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if (map.get(token[i]) - map.get(token[i - 1]) != 1) {
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isStraight = false;
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break;
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}
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}
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if (isStraight && token.length == 5) return CardType.Straight;
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return CardType.Sigal;
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}
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|
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private enum CardType {
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DoubleJoker, Bomb, Sigal, Double, Triple, Straight;
|
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}
|
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|
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public static void main(String[] args) {
|
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Main main = new Main();
|
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Scanner in = new Scanner(System.in);
|
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while (in.hasNextLine()) {
|
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String s = in.nextLine();
|
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String[] token = s.split("-");
|
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System.out.println(main.play(token[0], token[1]));
|
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}
|
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}
|
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}
|
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```
|
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|
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# 11. 去哪儿-二分查找
|
||
|
||
对于有重复元素的有序数组，二分查找需要注意以下要点：
|
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|
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- if (val <= A[m]) h = m;
|
||
- 因为 h 的赋值为 m 而不是 m - 1，因此 while 循环的条件也就为 l < h。（如果是 m - 1 循环条件为 l <= h）
|
||
|
||
```java
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public int getPos(int[] A, int n, int val) {
|
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int l = 0, h = n - 1;
|
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while (l < h) {
|
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int m = l + (h - l) / 2;
|
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if (val <= A[m]) h = m;
|
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else l = m + 1;
|
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}
|
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return A[h] == val ? h : -1;
|
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}
|
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```
|
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|
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# 12. 去哪儿-首个重复字符
|
||
|
||
```java
|
||
public char findFirstRepeat(String A, int n) {
|
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boolean[] hasAppear = new boolean[256];
|
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for (int i = 0; i < n; i++) {
|
||
char c = A.charAt(i);
|
||
if(hasAppear[c]) return c;
|
||
hasAppear[c] = true;
|
||
}
|
||
return ' ';
|
||
}
|
||
```
|
||
|
||
# 13. 去哪儿-寻找Coder
|
||
|
||
```java
|
||
public String[] findCoder(String[] A, int n) {
|
||
List<Pair<String, Integer>> list = new ArrayList<>();
|
||
for (String s : A) {
|
||
int cnt = 0;
|
||
String t = s.toLowerCase();
|
||
int idx = -1;
|
||
while (true) {
|
||
idx = t.indexOf("coder", idx + 1);
|
||
if (idx == -1) break;
|
||
cnt++;
|
||
}
|
||
if (cnt != 0) {
|
||
list.add(new Pair<>(s, cnt));
|
||
}
|
||
}
|
||
Collections.sort(list, (o1, o2) -> (o2.getValue() - o1.getValue()));
|
||
String[] ret = new String[list.size()];
|
||
for (int i = 0; i < list.size(); i++) {
|
||
ret[i] = list.get(i).getKey();
|
||
}
|
||
return ret;
|
||
}
|
||
|
||
// 牛客网无法导入 javafx.util.Pair，这里就自己实现一下 Pair 类
|
||
private class Pair<T, K> {
|
||
T t;
|
||
K k;
|
||
|
||
Pair(T t, K k) {
|
||
this.t = t;
|
||
this.k = k;
|
||
}
|
||
|
||
T getKey() {
|
||
return t;
|
||
}
|
||
|
||
K getValue() {
|
||
return k;
|
||
}
|
||
}
|
||
```
|
||
|
||
# 14. 美团-最大差值
|
||
|
||
贪心策略。
|
||
|
||
```java
|
||
public int getDis(int[] A, int n) {
|
||
int max = 0;
|
||
int soFarMin = A[0];
|
||
for (int i = 1; i < n; i++) {
|
||
if(soFarMin > A[i]) soFarMin = A[i];
|
||
else max = Math.max(max, A[i]- soFarMin);
|
||
}
|
||
return max;
|
||
}
|
||
```
|
||
|
||
# 15. 美团-棋子翻转
|
||
|
||
```java
|
||
public int[][] flipChess(int[][] A, int[][] f) {
|
||
int[][] direction = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
|
||
for (int[] ff : f) {
|
||
for (int[] dd : direction) {
|
||
int r = ff[0] + dd[0] - 1, c = ff[1] + dd[1] - 1;
|
||
if(r < 0 || r > 3 || c < 0 || c > 3) continue;
|
||
A[r][c] ^= 1;
|
||
}
|
||
}
|
||
return A;
|
||
}
|
||
```
|
||
|
||
# 16. 美团-拜访
|
||
|
||
```java
|
||
private Set<String> paths;
|
||
private List<Integer> curPath;
|
||
|
||
public int countPath(int[][] map, int n, int m) {
|
||
paths = new HashSet<>();
|
||
curPath = new ArrayList<>();
|
||
for (int i = 0; i < n; i++) {
|
||
for (int j = 0; j < m; j++) {
|
||
if (map[i][j] == 1) {
|
||
map[i][j] = -1;
|
||
int[][] leftRightDirection = {{1, 0}, {-1, 0}};
|
||
int[][] topDownDirection = {{0, 1}, {0, -1}};
|
||
for (int[] lr : leftRightDirection) {
|
||
for (int[] td : topDownDirection) {
|
||
int[][] directions = {lr, td};
|
||
backtracking(map, n, m, i, j, directions);
|
||
}
|
||
}
|
||
return paths.size();
|
||
}
|
||
}
|
||
}
|
||
return 0;
|
||
}
|
||
|
||
private void backtracking(int[][] map, int n, int m, int r, int c, int[][] directions) {
|
||
if (map[r][c] == 2) {
|
||
String path = "";
|
||
for (int num : curPath) {
|
||
path += num;
|
||
}
|
||
paths.add(path);
|
||
return;
|
||
}
|
||
for (int i = 0; i < directions.length; i++) {
|
||
int nextR = r + directions[i][0];
|
||
int nextC = c + directions[i][1];
|
||
if (nextR < 0 || nextR >= n || nextC < 0 || nextC >= m || map[nextR][nextC] == -1) continue;
|
||
map[nextR][nextC] = map[nextR][nextC] == 2 ? 2 : -1;
|
||
curPath.add(nextR);
|
||
curPath.add(nextC);
|
||
backtracking(map, n, m, nextR, nextC, directions);
|
||
curPath.remove(curPath.size() - 1);
|
||
curPath.remove(curPath.size() - 1);
|
||
map[nextR][nextC] = map[nextR][nextC] == 2 ? 2 : 0;
|
||
}
|
||
}
|
||
```
|
||
|
||
# 17. 美团-直方图内最大矩形
|
||
|
||
```java
|
||
public int countArea(int[] A, int n) {
|
||
int max = 0;
|
||
for (int i = 0; i < n; i++) {
|
||
int min = A[i];
|
||
for (int j = i; j < n; j++) {
|
||
min = Math.min(min, A[j]);
|
||
max = Math.max(max, min * (j - i + 1));
|
||
}
|
||
}
|
||
return max;
|
||
}
|
||
```
|
||
|
||
# 18. 美团-字符串计数
|
||
|
||
字符串都是小写字符，可以把字符串当成是 26 进制。但是字典序的比较和普通的整数比较不同，是从左往右进行比较，例如 "ac" 和 "abc"，字典序的比较结果为 "ac" > "abc"，如果按照整数方法比较，因为 "abc" 是三位数，显然更大。
|
||
|
||
由于两个字符串的长度可能不想等，在 s1 空白部分和 s2 对应部分进行比较时，应该把 s1 的空白部分看成是 'a' 字符进行填充的。
|
||
|
||
还有一点要注意的是，s1 到 s2 长度为 len<sub>i</sub> 的字符串个数只比较前面 i 个字符。例如 'aaa' 和 'bbb' ，长度为 2 的个数为 'aa' 到 'bb' 的字符串个数，不需要考虑后面部分的字符。
|
||
|
||
在统计个数时，从 len1 开始一直遍历到最大合法长度，每次循环都统计长度为 i 的子字符串个数。
|
||
|
||
```java
|
||
String s1 = in.next();
|
||
String s2 = in.next();
|
||
int len1 = in.nextInt();
|
||
int len2 = in.nextInt();
|
||
int len = Math.min(s2.length(), len2);
|
||
int[] subtractArr = new int[len];
|
||
for (int i = 0; i < len; i++) {
|
||
char c1 = i < s1.length() ? s1.charAt(i) : 'a';
|
||
char c2 = s2.charAt(i);
|
||
subtractArr[i] = c2 - c1;
|
||
}
|
||
int ret = 0;
|
||
for (int i = len1; i <= len; i++) {
|
||
for (int j = 0; j < i; j++) {
|
||
ret += subtractArr[j] * Math.pow(26, i - j - 1);
|
||
}
|
||
}
|
||
System.out.println(ret - 1);
|
||
```
|
||
|
||
# 19. 美团-平均年龄
|
||
|
||
```java
|
||
int W = in.nextInt();
|
||
double Y = in.nextDouble();
|
||
double x = in.nextDouble();
|
||
int N = in.nextInt();
|
||
while (N-- > 0) {
|
||
Y++; // 老员工每年年龄都要加 1
|
||
Y += (21 - Y) * x;
|
||
}
|
||
System.out.println((int) Math.ceil(Y));
|
||
```
|
||
|
||
# 20. 百度-罪犯转移
|
||
|
||
部分和问题，将每次求的部分和缓存起来。
|
||
|
||
```java
|
||
int n = in.nextInt();
|
||
int t = in.nextInt();
|
||
int c = in.nextInt();
|
||
int[] values = new int[n];
|
||
for (int i = 0; i < n; i++) {
|
||
values[i] = in.nextInt();
|
||
}
|
||
int cnt = 0;
|
||
int totalValue = 0;
|
||
for (int s = 0, e = c - 1; e < n; s++, e++) {
|
||
if (s == 0) {
|
||
for (int j = 0; j < c; j++) totalValue += values[j];
|
||
} else {
|
||
totalValue = totalValue - values[s - 1] + values[e];
|
||
}
|
||
if (totalValue <= t) cnt++;
|
||
}
|
||
System.out.println(cnt);
|
||
```
|
||
|
||
# 22. 百度-裁减网格纸
|
||
|
||
```java
|
||
int n = in.nextInt();
|
||
int minX, minY, maxX, maxY;
|
||
minX = minY = Integer.MAX_VALUE;
|
||
maxX = maxY = Integer.MIN_VALUE;
|
||
for (int i = 0; i < n; i++) {
|
||
int x = in.nextInt();
|
||
int y = in.nextInt();
|
||
minX = Math.min(minX, x);
|
||
minY = Math.min(minY, y);
|
||
maxX = Math.max(maxX, x);
|
||
maxY = Math.max(maxY, y);
|
||
}
|
||
System.out.println((int) Math.pow(Math.max(maxX - minX, maxY - minY), 2));
|
||
```
|
||
|
||
# 23. 百度-钓鱼比赛
|
||
|
||
P ( 至少钓一条鱼 ) = 1 - P ( 一条也钓不到 )
|
||
|
||
坑：读取概率矩阵的时候，需要一行一行进行读取，而不能直接用 in.nextDouble()。
|
||
|
||
```java
|
||
public static void main(String[] args) {
|
||
Scanner in = new Scanner(System.in);
|
||
while (in.hasNext()) {
|
||
int n = in.nextInt();
|
||
int m = in.nextInt();
|
||
int x = in.nextInt();
|
||
int y = in.nextInt();
|
||
int t = in.nextInt();
|
||
in.nextLine(); // 坑
|
||
double pcc = 0.0;
|
||
double sum = 0.0;
|
||
for (int i = 1; i <= n; i++) {
|
||
String[] token = in.nextLine().split(" "); // 坑
|
||
for (int j = 1; j <= m; j++) {
|
||
double p = Double.parseDouble(token[j - 1]);
|
||
// double p = in.nextDouble();
|
||
sum += p;
|
||
if (i == x && j == y) {
|
||
pcc = p;
|
||
}
|
||
}
|
||
}
|
||
double pss = sum / (n * m);
|
||
pcc = computePOfIRT(pcc, t);
|
||
pss = computePOfIRT(pss, t);
|
||
System.out.println(pcc > pss ? "cc" : pss > pcc ? "ss" : "equal");
|
||
System.out.printf("%.2f\n", Math.max(pcc, pss));
|
||
}
|
||
}
|
||
|
||
// compute probability of independent repeated trials
|
||
private static double computePOfIRT(double p, int t) {
|
||
return 1 - Math.pow((1 - p), t);
|
||
}
|
||
```
|
||
|
||
# 24. 百度-蘑菇阵
|
||
|
||
这题用回溯会超时，需要用 DP。
|
||
|
||
dp[i][j] 表示到达 (i,j) 位置不会触碰蘑菇的概率。对于 N\*M 矩阵，如果 i == N || j == M，那么 (i,j) 只能有一个移动方向；其它情况下能有两个移动方向。
|
||
|
||
考虑以下矩阵，其中第 3 行和第 3 列只能往一个方向移动，而其它位置可以有两个方向移动。
|
||
|
||
|
||
```java
|
||
int N = in.nextInt();
|
||
int M = in.nextInt();
|
||
int K = in.nextInt();
|
||
boolean[][] mushroom = new boolean[N][M];
|
||
while (K-- > 0) {
|
||
int x = in.nextInt();
|
||
int y = in.nextInt();
|
||
mushroom[x - 1][y - 1] = true;
|
||
}
|
||
double[][] dp = new double[N][M];
|
||
dp[0][0] = 1;
|
||
for (int i = 0; i < N; i++) {
|
||
for (int j = 0; j < M; j++) {
|
||
if (mushroom[i][j]) dp[i][j] = 0;
|
||
else {
|
||
double cur = dp[i][j];
|
||
if (i == N - 1 && j == M - 1) break;
|
||
if (i == N - 1) dp[i][j + 1] += cur;
|
||
else if (j == M - 1) dp[i + 1][j] += cur;
|
||
else {
|
||
dp[i][j + 1] += cur / 2;
|
||
dp[i + 1][j] += cur / 2;
|
||
}
|
||
}
|
||
}
|
||
}
|
||
System.out.printf("%.2f\n", dp[N - 1][M - 1]);
|
||
```
|