HDU 2830 Matrix Swapping II

Matrix Swapping II

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1260    Accepted Submission(s): 839

Problem Description

Given an N * M matrix with each entry equal to 0 or 1. We can find some rectangles in the matrix whose entries are all 1, and we define the maximum area of such rectangle as this matrix’s goodness. 

We can swap any two columns any times, and we are to make the goodness of the matrix as large as possible.

 

Input

There are several test cases in the input. The first line of each test case contains two integers N and M (1 ≤ N,M ≤ 1000). Then N lines follow, each contains M numbers (0 or 1), indicating the N * M matrix

 

Output

Output one line for each test case, indicating the maximum possible goodness.

 

Sample Input

3 4 1011 1001 0001 3 4 1010 1001 0001

 

Sample Output

4 2 Note: Huge Input, scanf() is recommended.

 

Source

2009 Multi-University Training Contest 2 - Host by TJU

 

解題思路：因爲列是可以任意交換的，記錄下到當前行爲止列的最大的連續1的個數，每行排個序，這樣就變成了以當前列爲高度的矩形的面積了，然後掃過去更新最大值即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define Max 1005
using namespace std;
char matrix[Max][Max];
bool cmp(int a,int b)
{return a>b;}
int main()
{
    int m,n,i,j,h[Max],sum[Max],ans;
    while(~scanf("%d%d",&m,&n))
    {
        ans=0;
        memset(matrix,0,sizeof(matrix));
        memset(h,0,sizeof(h));
        memset(sum,0,sizeof(sum));
        getchar();
        for(i=1;i<=m;i++)
        {
            for(j=1;j<=n;j++)
            {
                scanf("%c",&matrix[i][j]);
                h[j]=matrix[i][j]=='0'?0:h[j]+1;
                sum[j]=h[j];
            }
            getchar();
            sort(sum+1,sum+1+n,cmp);
            for(j=1;j<=n;j++)
                ans=max(ans,sum[j]*j);
            //cout<<ans<<endl;
        }
        printf("%d\n",ans);
    }
    return 0;
}