洛谷1387（基础二维dp）

题目很简单，数据也很小，但是思路不妨借鉴：dp[i][j]代表以（i，j）为右下角的最长正方形边长。

类比一维里面设“以XX为结尾的最XXX（所求）”。

另外define不要乱用！尤其这种min套min，debug两行泪。

 #include <cstdio>
 #include <algorithm>
 using namespace std;

 int n, m, ans, a, dp[][];

 int main() {
     scanf("%d %d", &n, &m);
     for (int i = ; i <= n; i++) {
         for (int j = ; j <= m; j++) {
             scanf("%d", &a);
             if (a)    dp[i][j] = min(dp[i - ][j - ], min(dp[i - ][j], dp[i][j - ])) + ;
             ans = max(ans, dp[i][j]);
         }
     }
     printf("%d\n", ans);
     return ;
 }

当然也可以无脑暴力乱搞了，二维前缀和+二分：

 #include <cstdio>
 #define min(a, b)   a < b ? a : b

 int n, m, ans, a[][], cnt[][];

 int main() {
     scanf("%d %d", &n, &m);
     for (int i = ; i <= n; ++i)
         for (int j = ; j <= m; ++j)
             scanf("%d", &a[i][j]);
     for (int i = ; i <= n; ++i)
         for (int j = ; j <= m; ++j)
             cnt[i][j] = cnt[i - ][j] + cnt[i][j - ] - cnt[i - ][j - ] + (a[i][j] == );

     int l = , r = min(n, m);
     auto ok = [](int len) {
         for (int i = ; i + len -  <= n; ++i)
             for (int j = ; j + len -  <= m; ++j)
                 if (cnt[i + len - ][j + len - ] - cnt[i + len - ][j - ] - cnt[i - ][j + len - ] + cnt[i - ][j - ] == )
                     return true;
         return false;
     };
     while (l <= r) {
         int mid = (l + r) >> ;
         if (ok(mid)) {
             ans = mid, l = mid + ;
         } else  r = mid - ;
     }
     printf("%d\n", ans);
     return ;
 }