CF 375B Maximum Submatrix 2[预处理计数排序]

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转载最新推荐文章于 2020-08-19 20:38:15 发布 · 171 阅读

本文介绍了一种算法，用于解决给定由0和1组成的矩阵中，通过行交换操作后能形成的全1子矩阵的最大面积问题。该算法首先进行预处理以计算每一行连续1的数量，然后对于每一列，利用计数排序对这些数量进行排序，并计算可能的最大全1子矩阵面积。

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.

Examples

input

1 1
1

output

input

2 2
10
11

output

input

output

交换行，求全1子矩阵最大


DP预处理a[i][j]  i行j列到右面有几个连续的1
枚举从那一列开始，按a来给行排序，统计就行了
排序可以用计数排序，然而时间没有明显改善

//
//  main.cpp
//  cf375b
//
//  Created by Candy on 9/15/16.
//  Copyright © 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int N=5005;
int n,m,a[N][N],ans=0;
char s[N];
int t[N],h[N],ne[N];

int cou[N],srt[N];
void buc(){
    memset(cou,0,sizeof(cou));
    int v=n;
    for(int i=1;i<=n;i++) cou[t[i]]++;
    for(int i=1;i<=v;i++) cou[i]+=cou[i-1];
    for(int i=n;i>=1;i--){
        srt[cou[t[i]]--]=t[i];
    }
}
int sol(int j){
    int ans=0;
    for(int i=1;i<=n;i++) t[i]=a[i][j];
    //sort(t+1,t+1+n);
    buc();
    for(int i=n;i>=1;i--)
        ans=max(ans,(n-i+1)*srt[i]);
    return ans;
}
int main(int argc, const char * argv[]) {
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%s",s);
        for(int j=m-1;j>=0;j--){
            if(s[j]=='1') a[i][j+1]=a[i][j+2]+1;
            else a[i][j+1]=0;
        }
    }
    
    for(int j=1;j<=m;j++)
        ans=max(ans,sol(j));
    printf("%d",ans);
    return 0;
}