B. Maximum Submatrix 2----dp

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 11

output

1

input

2 21011

output

2

input

4 3100011000101

output

2

题目链接：http://codeforces.com/contest/375/problem/B

这个题本以为是暴力或者是利用某些奇技淫巧的很优雅的暴力，但是没有想到是个dp。。

真的好难想到这个题是dp，看了卿学姐的博客

http://www.cnblogs.com/qscqesze/p/4607185.html

代码：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define LL long longusing namespace std;char s[5555][5555];int dp[5555][5555];bool cmp(int a,int b){    return a>b;}int main(){    int n,m;    scanf("%d%d",&n,&m);    int ans=0;    for(int i=1;i<=n;i++){        scanf("%s",s[i]+1);    }    for(int i=1;i<=n;i++){        for(int j=1;j<=m;j++){            if(s[i][j]=='1'){                dp[j][i]=dp[j-1][i]+1;            }        }    }    for(int i=1;i<=m;i++){        sort(dp[i]+1,dp[i]+1+n,cmp);        for(int j=1;j<=n;j++){            if(dp[i][j]==0){                 break;            }            ans=max(dp[i][j]*j,ans);        }    }    cout<<ans<<endl;}