python/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/6_0.cpp

#include <bits/stdc++.h>
/*
测试用例I：
5 6
1 0 1 0 0 0
0 1 0 1 0 0
1 0 0 0 1 0
0 1 0 0 0 1
1 0 1 0 0 0

答案
4

测试用例II：
5 6
1 0 1 0 0 0
0 1 0 1 1 0
1 0 0 0 1 0
0 1 0 0 0 1
1 0 1 0 0 0

答案
3
*/
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 110;
int a[N][N];
int res;

int main() {
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            cin >> a[i][j];

    // 1e8 =100000000
    //     1500000000
    for (int i = 1; i <= n; i++)                   // 1000
        for (int j = 1; j <= m; j++) {             // 1000
            for (int k = 0;; k++) {                // 1000 *SQRT(2)=1500
                if (i + k > n || j + k > m) break; //越界
                if (a[i + k][j + k] == 0) break;
                //从[i,j]出发，边长长度为k+1的矩形，内部数字为1的有多少个

                res = max(res, k + 1);
            }
        }
    cout << res << endl;
    return 0;
}