CF 375B Maximum Submatrix 2[預處理 計數排序]

Candy?發表於2016-09-15
排序
B. Maximum Submatrix 2
time limit per test
2 seconds
memory limit per test
512 megabytes
input
standard input
output
standard output

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
1
input
2 2
10
11
output
2
input
4 3
100
011
000
101
output
2
交換行，求全1子矩陣最大

DP預處理a[i][j]  i行j列到右面有幾個連續的1
列舉從那一列開始，按a來給行排序，統計就行了
排序可以用計數排序，然而時間沒有明顯改善
//
//  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;
}

 

 

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