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HDU 2830 Matrix Swapping II

2024-07-14 22:12:07   221人阅读

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.
 
做法参考网上，记录每一行的连续个数，排序后dp即可。盗张图

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
using namespace std;
int n,m;
char str[1100];
int num[1100],sum[1100];
int cmp(int a,int b)
{
    return a>b;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        getchar();
        char ch;
        int ans=0;
        memset(num,0,sizeof(num));
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                scanf("%c",&ch);
                if(ch=='1')
                    num[j]++;
                else
                    num[j]=0;
                sum[j]=num[j];
            }
            sort(sum+1,sum+m+1,cmp);
            for(int i=1;i<=m;i++)
                 ans=max(ans,sum[i]*i);
            getchar();
        }
        printf("%d\n",ans);
    }
    return 0;
}



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