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): 793    Accepted Submission(s): 537

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

 

 

Recommend

gaojie

 

 

 

对每一行进行处理然后再叠加，到每一行用num[i]记下到这一行有多少个1

 

例如：

 

1 0 1 1          num[i]的记录就是： 1 0 1 1                    

 

1 0 0 1                             2 0 0 2         

 

0 0 0 1                             0 0 0 3

 

然后对num[i]从大到小排一下，求出读取到当前行能够得到的最大面积。

 

也就是最右边放着最多的1，依次类推。那么当前最大的面积就是 MAX（max,num[i]*i）了

 

 

 

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

using namespace std;

const int maxn=1010;

int num[maxn],row[maxn];

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

int main(){

    //freopen("input.txt","r",stdin);

    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        getchar();
        memset(num,0,sizeof(num));
        int i,j,ans=0;
        char ch;
        for(i=1;i<=n;i++){
            for(j=1;j<=m;j++){
                scanf("%c",&ch);
                if(ch=='1')
                    num[j]++;
                else
                    num[j]=0;
                row[j]=num[j];
            }
            sort(row+1,row+m+1,cmp);
            for(j=1;j<=m;j++)
                if(ans<row[j]*j)
                    ans=row[j]*j;
            getchar();
        }
        printf("%d\n",ans);
    }
    return 0;
}