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 = 2550;
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 = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            scanf("%d", &a[i][j]);
        }
    }

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            if (!a[i][j]) {
                s1[i][j] = s1[i-1][j] + 1;//up
                s2[i][j] = s2[i][j-1] + 1;//left
            } else {
                dp[i][j] = min(dp[i-1][j-1], min(s1[i-1][j], s2[i][j-1])) + 1;
                ans = max(ans, dp[i][j]);
            }
        }
    }
    memset(dp, 0, sizeof dp);
    memset(s2, 0, sizeof s2);
    for (int i = 1; i <= n; i++) {
        for (int j = m; j; j--) {
            if (!a[i][j])    s2[i][j] = s2[i][j+1] + 1;//right
            else {
                dp[i][j] = min(dp[i-1][j+1], min(s1[i-1][j], s2[i][j+1])) + 1;
                ans = max(ans, dp[i][j]);
            }
        }
    }

    printf("%d
", ans);
    return 0;
}

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[2550][2550], b[2550][2550], sum1[2550][2550], sum2[2550][2550], dp[2][2550];

void DP(int a[][2550], int sum[][2550]) {
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            if (a[i][j]) {
                auto ok = [](int i, int j, int tmp, int sum[][2550]) {
                    return  sum[i][j] - sum[i - tmp][j] - sum[i][j - tmp] + sum[i - tmp][j - tmp] == tmp;
                };
                int l = 1, r = dp[(i - 1)&1][j - 1] + 1, t;
                while (l <= r) {
                    int mid = (l + r) >> 1;
                    if (ok(i, j, mid, sum)) {
                        t = mid;
                        l = mid + 1;
                    } else  r = mid - 1;
                }
                dp[i&1][j] = t;
            } else  dp[i&1][j] = 0;
            ans = max(ans, dp[i&1][j]);
        }
    }
}

int main() {
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++) {
            scanf("%d", &a[i][j]);
            b[i][m - j + 1] = a[i][j];
        }
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++) {
            sum1[i][j] = sum1[i - 1][j] + sum1[i][j - 1] - sum1[i - 1][j - 1] + a[i][j];
            sum2[i][j] = sum2[i - 1][j] + sum2[i][j - 1] - sum2[i - 1][j - 1] + b[i][j];
        }
    DP(a, sum1);
    memset(dp, 0, sizeof dp);
    DP(b, sum2);
    printf("%d
", ans);
    return 0;
}