python/TangDou/LanQiaoBei/LanQiaoBei13ShengSai_ZJ_II/6_3.cpp

#include <bits/stdc++.h>
/*
测试用例：
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
2
*/
using namespace std;
const int N = 110;
int a[N][N];
int s[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];
            //维护二维前缀和
            s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i][j];
        }

    //直接在原数组上递推
    for (int i = n - 1; i >= 1; i--)        //从倒数第二行开始
        for (int j = 1; j <= m; j++)        //这个需要考虑一下边界问题
            if (a[i][j]) {                  //只有当前位置大于0才有考虑的价值
                a[i][j] += a[i + 1][j + 1]; //递推式,当前位置可以向下扩展的最长对角线长度

                //利用二维前缀和计算区域的总和
                int x1 = i, y1 = j, x2 = i + a[i][j] - 1, y2 = j + a[i][j] - 1;
                int cnt = s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1];
                if (cnt == a[i][j])
                    res = max(res, a[i][j]); //在递推过程中不断求最大值
            }
    //输出
    cout << res << endl;
    return 0;
}