mirror of
https://github.com/huihut/interview.git
synced 2024-03-22 13:10:48 +08:00
103 lines
2.2 KiB
C++
103 lines
2.2 KiB
C++
#include <iostream>
|
||
#include <math.h>
|
||
|
||
using namespace std;
|
||
|
||
// 循环赛日程安排函数声明
|
||
void MatchTable(int k, int n, int **table);
|
||
|
||
int main()
|
||
{
|
||
int n = 0, k = 0;
|
||
|
||
// 用户界面
|
||
cout << "---------------- 循环赛日程安排问题 ----------------" << endl;
|
||
cout << "请输入k(k>=0),构成 n=(2^k) 个选手的循环赛" << endl;
|
||
|
||
// 输入k值
|
||
cin >> k;
|
||
|
||
// 判断输入数据合法性,包括检查输入是否为数字,k值是否大于0
|
||
if (cin.fail() || k < 0)
|
||
{
|
||
cout << "输入k错误!" << endl;
|
||
system("pause");
|
||
return 0;
|
||
}
|
||
|
||
// 计算比赛日程表大小
|
||
n = pow(2, k);
|
||
|
||
// 分配日程表空间
|
||
int **table = new int *[n + 1];
|
||
for (int i = 0; i <= n; i++)
|
||
{
|
||
table[i] = new int[n + 1];
|
||
}
|
||
|
||
// 进行循环赛日程安排,生成日程表
|
||
MatchTable(k, n, table);
|
||
|
||
// 显示输出
|
||
cout << "------------------------------------------------" << endl;
|
||
for (int i = 1; i <= n; i++)
|
||
{
|
||
for (int j = 1; j <= n; j++)
|
||
{
|
||
cout << table[i][j] << "\t";
|
||
}
|
||
cout << endl;
|
||
}
|
||
cout << "------------------------------------------------" << endl;
|
||
|
||
// 暂停查看结果
|
||
system("pause");
|
||
|
||
// 释放内存
|
||
for (int i = 0; i <= n; i++)
|
||
delete[] table[i];
|
||
delete[] table;
|
||
|
||
// 指针置空
|
||
table = NULL;
|
||
|
||
return 0;
|
||
}
|
||
|
||
// 进行循环赛日程安排,生成日程表
|
||
void MatchTable(int k, int n, int **table)
|
||
{
|
||
// 设置日程表第一行的值
|
||
for (int i = 1; i <= n; i++)
|
||
table[1][i] = i;
|
||
|
||
// 每次填充的起始填充位置
|
||
int begin = 1;
|
||
|
||
// 用分治法分separate份,循环求解
|
||
for (int separate = 1; separate <= k; separate++)
|
||
{
|
||
// 日程表进行划分
|
||
n /= 2;
|
||
|
||
// flag为每一小份的列的标记
|
||
for (int flag = 1; flag <= n; flag++)
|
||
{
|
||
// 操作行
|
||
for (int i = begin + 1; i <= 2 * begin; i++)
|
||
{
|
||
// 操作列
|
||
for (int j = begin + 1; j <= 2 * begin; j++)
|
||
{
|
||
// 把左上角的值赋给右下角
|
||
table[i][j + (flag - 1) * begin * 2] = table[i - begin][j + (flag - 1) * begin * 2 - begin];
|
||
// 把右上角的值赋给左下角
|
||
table[i][j + (flag - 1) * begin * 2 - begin] = table[i - begin][j + (flag - 1) * begin * 2];
|
||
}
|
||
}
|
||
}
|
||
// 进入日程表的下一个划分进行填充
|
||
begin *= 2;
|
||
}
|
||
}
|