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

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2025-08-30 18:35:01 +08:00
#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;
}