AcWing 5415. 仓库规划
原题链接
简单
作者:
拼凑回忆
,
2024-01-23 15:37:51
,
所有人可见
,
阅读 33
#include <iostream>
using namespace std;
const int N = 1010;
int a[N][N], f[N];
int n, m;
int compare(int x, int y)
{
int k;
for (k = 0; k < m; k ++ )
{
if (a[x][k] >= a[y][k]) return 0;
}
return 1;
}
int main()
{
cin >> n >> m;
for (int i = 0; i < n; i ++ )
for (int j = 0; j < m; j ++ )
cin >> a[i][j];
for (int i = 0; i < n; i ++ )
{
for(int j = 0; j < n; j ++)
{
//如果和本身比较则跳过
if (i == j) continue;
//比较i 和j 如果 a[i] < a[j] 的全部则进入判断
if (compare(i, j))
{
//取编号最小的
if (f[i] == 0 || f[i] > j + 1) f[i] = j + 1;
}
}
}
for (int i = 0; i < n; i ++ ) cout << f[i] << endl;
}