開始沒看到要求對角線以外的地方不能是0，以為是個xx題。。。照著題解思路寫的，很妙啊

題意：給定01矩陣
　　求矩陣中最長的只有對角線是1的正方形的對角線長度

x[i][j]從(i,j)向左/右（不包括(i,j))的連續的0的數量

y[i][j]從(i,j)向上（不包括(i,j))的連續的0的數量

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const int eps = 1e-8;
 5 typedef long double LD;
 6 typedef pair<int
 
, int> pii;
 7 #define lop(i,a,b) for(register int i = (a); i <= (b); ++i)
 8 #define dlop(i,a,b) for(register int i = (a); i >= (b); --i)
 9 #define bl(i) for(int i = head[u]; i; i = G[i].next)
10 //char buf[50<<20], *p1=buf;
11 //#define getchar() (*p1++)
12 #define make_pair MP
13 inline int
 
 read(){
14   register int c = getchar(), x = 0, f = 1;
15   while(!isdigit(c)){if (c == ‘-‘) f = -1; c = getchar();}
16   while(isdigit(c)) x = x*10+(c&15), c = getchar();
17   return x * f;
18 }
19 int n, m, s, dp[2503][2505], x[2503][2505], y[2503][2505];
20 bool matrix[2503][2505];
21 int main(void
 
){
22   //freopen("data.in", "r", stdin);
23   //fread(buf,1,50<<20,stdin);
24   int ans = 0;
25   n = read(), m = read();
26   lop(i,1,n) lop(j,1,m) {
27      matrix[i][j] = read();
28      if (matrix[i][j]) dp[i][j] = min(dp[i-1][j-1], min(x[i][j-1], y[i-1][j]))+1, ans = max(ans, dp[i][j]);
29      else x[i][j] = x[i][j-1] + 1, y[i][j] = y[i-1][j] + 1;
30   }
31   memset(x,0,sizeof(x));
32   memset(dp,0,sizeof(dp));
33   lop(i,1,n) dlop(j,m,1) {
34     if (matrix[i][j]) dp[i][j] = min(dp[i-1][j+1], min(x[i][j+1], y[i-1][j]))+1, ans = max(ans, dp[i][j]);
35     else x[i][j] = x[i][j+1] + 1;
36   }
37   cout << ans;
38   return 0;
39 }

[P1736]創意吃魚法[DP]