【ZJOI2007】棋盘制作

Description

给定一个01矩阵，求出最大的正方形子矩阵和最大长方形子矩阵，并保证取出的矩阵中的元素都为0/1

Solution

为了使题目变成上述表述的形式，我们首先将读入的矩阵的部分元素xor1，使其变成一般形式

之后，这道题变成了两道题的合体：洛谷P2701巨大的牛棚与P4147玉蟾宫

我们分开讨论，对于最大正方形，我们设计一个dp，定义f[i][j]表示以(i, j)为右下角的最大正方形的边长，状态转移方程为

f[i][j] = min(f[i - 1][j - 1], f[i - 1][j], f[i][j - 1]) + 1

那么答案也是显然的，ans = max(f[i][j] * f[i][j])

对于最大矩形，我们使用单调栈求解，具体步骤请见这里

注意，本题要求求出最大正方形与矩形，所以我们要将这两个算法对0/1矩阵各做一遍，时间复杂度为O(n^2)

Code

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int n, m;
int a[2010][2010], vis[2010][2010];
int ans1, f1[2010][2010], f2[2010][2010], ans2, b[2010]; 
int s[2010], p, w[2010];
int main() { 
    scanf("%d%d", &n, &m);
    for (register int i = 1; i <= m; ++i) vis[1][i] = vis[1][i - 1] ^ 1;
    for (register int i = 2; i <= n; ++i)
        for (register int j = 1; j <= m; j++)
            vis[i][j] = vis[i - 1][j] ^ 1;
    for (register int i = 1; i <= n; ++i)
        for (register int j = 1; j <= m; ++j) {
            scanf("%d", &a[i][j]);
            if(!vis[i][j]) a[i][j] ^= 1;
        }
    for (register int i = 1; i <= n; ++i)
        for (register int j = 1; j <= m; ++j)
            if (a[i][j]) {
                f1[i][j] = min(f1[i - 1][j - 1], min(f1[i - 1][j], f1[i][j - 1])) + 1;
                ans1 = max(ans1, f1[i][j]);
            }
    for (register int i = 1; i <= n; ++i)
        for (register int j = 1; j <= m; ++j)
            if (!a[i][j]) {
                f2[i][j] = min(f2[i - 1][j - 1], min(f2[i - 1][j], f2[i][j - 1])) + 1;
                ans1 = max(ans1, f2[i][j]);
            }
    printf("%d
", ans1 * ans1);
    for (register int i = 1; i <= n; ++i) {
        for (register int j = 1; j <= m; ++j)
            b[j] = a[i][j] ? b[j] + 1 : 0;
        p = 0;
        for (register int j = 1; j <= m + 1; ++j) {
            if (b[j] > s[p]) s[++p] = b[j], w[p] = 1;
            else {
                int tot = 0;
                while (b[j] < s[p]) {
                    tot += w[p];
                    ans2 = max(ans2, s[p] * tot);
                    p--;
                }
                s[++p] = b[j];
                w[p] = tot + 1;
            }
        }
    }
    for (register int i = 1; i <= n; ++i) {
        for (register int j = 1; j <= m; ++j)
            b[j] = a[i][j] == 0 ? b[j] + 1 : 0;
        p = 0;
        for (register int j = 1; j <= m + 1; ++j) {
            if (b[j] > s[p]) s[++p] = b[j], w[p] = 1;
            else {
                int tot = 0;
                while (b[j] < s[p]) {
                    tot += w[p];
                    ans2 = max(ans2, s[p] * tot);
                    p--;
                }
                s[++p] = b[j];
                w[p] = tot + 1;
            }
        }
    }
    printf("%d
", ans2);
    return 0;
}