poj3494 Largest Submatrix of All 1’s

最新推荐文章于 2025-08-15 16:44:13 发布

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原文链接：http://www.cnblogs.com/EdSheeran/p/9396510.html

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#数据结构与算法

本文探讨了如何寻找给定的m-by-n (0,1)-矩阵中最大的全1子矩阵，通过使用单调队列和前缀和数组来解决底部不一致的问题，提供了一个高效的算法实现。

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Largest Submatrix of All 1’s

链接:http://poj.org/problem?id=3494

Time Limit: 5000MS		Memory Limit: 131072K

Case Time Limit: 2000MS

Description

Given a m-by-n (0,1)-matrix, of all its submatrices of all 1’s which is the largest? By largest we mean that the submatrix has the most elements.

Input

The input contains multiple test cases. Each test case begins with m and n (1 ≤ m, n ≤ 2000) on line. Then come the elements of a (0,1)-matrix in row-major order on m lines each with n numbers. The input ends once EOF is met.

Output

For each test case, output one line containing the number of elements of the largest submatrix of all 1’s. If the given matrix is of all 0’s, output 0.

Sample Input

Sample Output

0
4
题意：最大全1子矩阵
题解：如果他们的底部在同一直线上，就是一道经典的单调队列：
底部不一样，就枚举底部好了，高度用一个前缀和数组维护；

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int M = 2005;
int mp[M][M], h[M];
struct Maxtri{
    int h, w;
}s[M];
int main(){
    int n, m;
    while(scanf("%d%d", &n, &m) != EOF){
        int ans = 0;
        memset(h, 0, sizeof(h));
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= m; j++)
                scanf("%d", &mp[i][j]);
        for(int i = 1; i <= n; i++){
            int tp = 0;
            for(int j = 1; j <= m; j++){
                int nw = 0;
                mp[i][j] ? h[j]++ : h[j] = 0;
                while(s[tp].h >= h[j] && tp){
                    nw += s[tp].w;
                    ans = max(ans, nw*s[tp].h);
                    tp--;
                }
                s[++tp].h = h[j];s[tp].w = nw+1;
            //    printf("%d %d %d\n", i, j, ans);
            }
            int nw = 0;
            while(tp){
                nw += s[tp].w;
                ans = max(ans, nw*s[tp--].h);
            }
        }
        printf("%d\n",ans);
    }
}