信奥一本通1122：计算鞍点

时间复杂度应该不会超的 不断枚举吧 sum是统计符合个数 如果没有就输出 not found

#include<bits/stdc++.h>

using namespace std;
int a[10][10]; 

int main(){
	int sum = 0;
	for(int i = 1; i <= 5; i++){
		for(int j =1; j <= 5; j++){
			cin >> a[i][j];
		}
	}
	for(int i = 1; i <= 5; i++){
		for(int j =1; j <= 5; j++){
			int maxn = -1e9;
			int minn = 1e9;
			int d = a[i][j];
			for(int k = 1; k <= 5; k++){
				maxn = max(maxn, a[i][k]);
			}
			for(int k = 1; k <= 5; k++){
				minn = min(minn, a[k][j]);
			}
			if(maxn == d && minn == d){
				sum++;
				cout << i << " " << j << " " << a[i][j];
			}
		}
	}
	if(sum ==0) cout << "not found";

    return 0;
}