洛谷1736（二维dp+预处理）

洛谷1387的进阶版，但很像。

1387要求是“全为1的正方形”，取dp[i][j] = min(dp[i-1][j-1], min(dp[i-1][j], dp[i][j-1]))吧？这个有“只有对角线可以有1”的要求，取的是dp[i][j] = min(dp[i-1][j-1], min(s1[i-1][j], s2[i][j-1]))，s1s2是预处理的两个数组，表示上方和左方有多少连续的0.另外本题左右方向对角线都算，所以得算两遍。

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

 const int maxn = ;
 int n, m, ans, a[maxn][maxn];
 int s1[maxn][maxn], s2[maxn][maxn], dp[maxn][maxn];

 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++) {
             if (!a[i][j]) {
                 s1[i][j] = s1[i-][j] + ;//up
                 s2[i][j] = s2[i][j-] + ;//left
             } else {
                 dp[i][j] = min(dp[i-][j-], min(s1[i-][j], s2[i][j-])) + ;
                 ans = max(ans, dp[i][j]);
             }
         }
     }
     memset(dp, , sizeof dp);
     memset(s2, , sizeof s2);
     for (int i = ; i <= n; i++) {
         for (int j = m; j; j--) {
             if (!a[i][j])    s2[i][j] = s2[i][j+] + ;//right
             else {
                 dp[i][j] = min(dp[i-][j+], min(s1[i-][j], s2[i][j+])) + ;
                 ans = max(ans, dp[i][j]);
             }
         }
     }

     printf("%d\n", ans);
     return ;
 }

而我自己做时……我这种满脑子二分前缀和的暴力分子大概没救了吧～不过这两种做法时间和空间上相差不多，反而是暴力快一点（逃

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

 int n, m, ans;
 int a[][], b[][], sum1[][], sum2[][], dp[][];

 void DP(int a[][], int sum[][]) {
     for (int i = ; i <= n; i++) {
         for (int j = ; j <= m; j++) {
             if (a[i][j]) {
                 auto ok = [](int i, int j, int tmp, int sum[][]) {
                     return  sum[i][j] - sum[i - tmp][j] - sum[i][j - tmp] + sum[i - tmp][j - tmp] == tmp;
                 };
                 int l = , r = dp[(i - )&][j - ] + , t;
                 while (l <= r) {
                     int mid = (l + r) >> ;
                     if (ok(i, j, mid, sum)) {
                         t = mid;
                         l = mid + ;
                     } else  r = mid - ;
                 }
                 dp[i&][j] = t;
             } else  dp[i&][j] = ;
             ans = max(ans, dp[i&][j]);
         }
     }
 }

 int main() {
     scanf("%d %d", &n, &m);
     for (int i = ; i <= n; i++)
         for (int j = ; j <= m; j++) {
             scanf("%d", &a[i][j]);
             b[i][m - j + ] = a[i][j];
         }
     for (int i = ; i <= n; i++)
         for (int j = ; j <= m; j++) {
             sum1[i][j] = sum1[i - ][j] + sum1[i][j - ] - sum1[i - ][j - ] + a[i][j];
             sum2[i][j] = sum2[i - ][j] + sum2[i][j - ] - sum2[i - ][j - ] + b[i][j];
         }
     DP(a, sum1);
     memset(dp, , sizeof dp);
     DP(b, sum2);
     printf("%d\n", ans);
     return ;
 }