请教，50分思路哪里有错QAQ

如果a[i][j]=0，用x来存储向左0的个数，用y来存储向上0的个数；如等于1，则用num来存储以这一个点为右下角的满足条件的正方形矩阵对角线长度

f[i][j].num = min3(f[i][j-1].x,f[i-1][j].y,f[i-1][j-1].num) +1;

#include<iostream>
#include<cstdio>
#define min3(a,b,c) min(a,min(b,c))
using namespace std;
int n,m,a[2501][2501],ans=0;
struct node {
    int x,y,num;
} f[2501][2501];
int main()
{
    cin >> n >> m;
    for (int i=1; i<=n; i++)
        for (int j=1; j<=m; j++) cin >> a[i][j];
    for (int i=1; i<=n; i++)
        for (int j=1; j<=m; j++)
            if (!a[i][j]) {
                f[i][j].x = f[i][j-1].x + 1;
                f[i][j].y = f[i-1][j].y + 1;
            }
            else {
                f[i][j].num = min3(f[i][j-1].x,f[i-1][j].y,f[i-1][j-1].num) +1;
                ans = max(ans,f[i][j].num);
            }
    cout << ans << endl;
    return 0;
}