mirror of
https://github.com/Kiritow/OJ-Problems-Source.git
synced 2024-03-22 13:11:29 +08:00
80 lines
2.2 KiB
C++
80 lines
2.2 KiB
C++
#include <cstdio>
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#include <cstdlib>
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#include <cstring>
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#include <algorithm>
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using namespace std;
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/*==========================================================*\
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| 划分树
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\*==========================================================*/
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#define MAX_SIZE 100005
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int sorted[MAX_SIZE];//已经排好序的数据
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int toleft[25][MAX_SIZE];
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int tree[25][MAX_SIZE];
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void build_tree(int left, int right, int deep)
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{
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int i;
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if (left == right) return ;
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int mid = (left + right) >> 1;
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int same = mid - left + 1; //位于左子树的数据
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for (i = left; i <= right; ++i) {//计算放于左子树中与中位数相等的数字个数
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if (tree[deep][i] < sorted[mid]) {
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--same;
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}
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}
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int ls = left;
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int rs = mid + 1;
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for (i = left; i <= right; ++i) {
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int flag = 0;
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if ((tree[deep][i] < sorted[mid]) || (tree[deep][i] == sorted[mid] && same > 0)) {
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flag = 1;
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tree[deep + 1][ls++] = tree[deep][i];
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if (tree[deep][i] == sorted[mid])
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same--;
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} else {
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tree[deep + 1][rs++] = tree[deep][i];
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}
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toleft[deep][i] = toleft[deep][i - 1]+flag;
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}
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build_tree(left, mid, deep + 1);
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build_tree(mid + 1, right, deep + 1);
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}
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int query(int left, int right, int k, int L, int R, int deep)
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{
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if (left == right)
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return tree[deep][left];
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int mid = (L + R) >> 1;
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int x = toleft[deep][left - 1] - toleft[deep][L - 1];//位于left左边的放于左子树中的数字个数
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int y = toleft[deep][right] - toleft[deep][L - 1];//到right为止位于左子树的个数
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int ry = right - L - y;//到right右边为止位于右子树的数字个数
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int cnt = y - x;//[left,right]区间内放到左子树中的个数
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int rx = left - L - x;//left左边放在右子树中的数字个数
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if (cnt >= k) {
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//printf("sss %d %d %d\n", xx++, x, y);
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return query(L + x, L + y - 1, k, L, mid, deep + 1);
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}
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else {
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//printf("qqq %d %d %d\n", xx++, x, y);
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return query(mid + rx + 1, mid + 1 + ry, k - cnt, mid + 1, R, deep + 1);
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}
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}
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int main()
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{
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int m, n;
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int a, b, k;
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int i;
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while (scanf("%d%d", &m, &n) == 2) {
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for (i = 1; i <= m; ++i) {
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scanf("%d", &sorted[i]);
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tree[0][i] = sorted[i];
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}
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sort(sorted + 1, sorted + 1 + m);
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build_tree(1, m, 0);
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for (i = 0; i < n; ++i) {
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scanf("%d%d%d", &a, &b, &k);
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printf("%d\n", query(a, b, k, 1, m, 0));
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}
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}
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return 0;
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}
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