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[105][105];

int main() {
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            scanf("%d", &a);
            if (a)    dp[i][j] = min(dp[i - 1][j - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1;
            ans = max(ans, dp[i][j]);
        }
    }
    printf("%d\n", ans);
    return 0;
}

 

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[105][105], cnt[105][105];

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)
            cnt[i][j] = cnt[i - 1][j] + cnt[i][j - 1] - cnt[i - 1][j - 1] + (a[i][j] == 0);

    int l = 0, r = min(n, m);
    auto ok = [](int len) {
        for (int i = 1; i + len - 1 <= n; ++i)
            for (int j = 1; j + len - 1 <= m; ++j)
                if (cnt[i + len - 1][j + len - 1] - cnt[i + len - 1][j - 1] - cnt[i - 1][j + len - 1] + cnt[i - 1][j - 1] == 0)
                    return true;
        return false;
    };
    while (l <= r) {
        int mid = (l + r) >> 1;
        if (ok(mid)) {
            ans = mid, l = mid + 1;
        } else  r = mid - 1;
    }
    printf("%d\n", ans);
    return 0;
}