Matrix Swapping II - HDU 2830 dp

Matrix Swapping II

Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1237 Accepted Submission(s): 824



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.

题意：交换任意列后，使得1的矩阵面积最大。

思路：还是求最大矩阵面积，只不过需要排序，具体实现看代码。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char s[1010][1010];
int n,m,val[1010][1010],l[1010];
int main()
{ int i,j,k,ans;
  while(~scanf("%d%d",&n,&m))
  { for(i=1;i<=n;i++)
    { scanf("%s",s[i]+1);
      for(j=1;j<=m;j++)
       if(s[i][j]=='1')
        val[i][j]=val[i-1][j]+1;
       else
        val[i][j]=0;
    }
    ans=0;
    for(i=1;i<=n;i++)
    { for(j=1;j<=m;j++)
       l[j]=j;
      sort(val[i]+1,val[i]+1+m);
      val[i][0]=val[i][m+1]=-1;
      for(j=1;j<=n;j++)
       while(val[i][l[j]-1]>=val[i][j])
        l[j]=l[l[j]-1];
      for(j=1;j<=n;j++)
      ans=max(ans,val[i][j]*(m-l[j]+1));
    }
    printf("%d\n",ans);
  }
}