CodeForces 375 B. Maximum Submatrix 2

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最新推荐文章于 2020-09-30 18:00:37 发布

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分类专栏： DP 文章标签： codeforces

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本文介绍了一种通过重新排列矩阵的行来寻找由1组成的最大子矩阵面积的方法。使用特定算法，通过对输入矩阵进行处理并排序，最终找到该子矩阵的最大面积。

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写法似乎太暴力了。。。

You are given a matrix consisting of digits zero and one, its size is n × m. You are allowed to rearrange its rows. What is the maximum area of the submatrix that only consists of ones and can be obtained in the given problem by the described operations?

Let's assume that the rows of matrix a are numbered from 1 to n from top to bottom and the columns are numbered from 1 to m from left to right. A matrix cell on the intersection of the i-th row and the j-th column can be represented as (i, j). Formally, a submatrix of matrix ais a group of four integers d, u, l, r (1 ≤ d ≤ u ≤ n; 1 ≤ l ≤ r ≤ m). We will assume that the submatrix contains cells (i, j) (d ≤ i ≤ u; l ≤ j ≤ r). The area of the submatrix is the number of cells it contains.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 5000). Next n lines contain m characters each — matrix a. Matrix a only contains characters: "0" and "1". Note that the elements of the matrix follow without any spaces in the lines.

Output

Print a single integer — the area of the maximum obtained submatrix. If we cannot obtain a matrix of numbers one, print 0.

Sample test(s)

input
1 1
1

output
1

input
2 2
10
11

output
2

input
4 3
100
011
000
101

output
2

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

char map[5500][5500];
int h[5500][5500],n,m,ans=0;

bool cmp(int a,int b)
{
    return a>b;
}

int main()
{
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++) scanf("%s",map[i]);
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {
            if(map[i-1][j-1]=='1') h[j][i]=h[j-1][i]+1;
        }
    }
    for(int i=1;i<=m;i++)
    {
        sort(h[i]+1,h[i]+n+1,cmp);
        for(int j=1;j<=n;j++)
        {
            if(h[i][j]==0) break;
            ans=max(ans,h[i][j]*j);
        }
    }
    printf("%d\n",ans);
    return 0;
}