HDu 2830 Matrix Swapping II（dp）

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 41011100100013 4101010010001

 

Sample Output

42Note: Huge Input, scanf() is recommended.

 

Source

题意：一列一列的移动图，问最大的1组成的面积？

思路：每一行中，对高排序，高就是a【i】，长就是i，我代码中是j,然后就可以求最优解了

</pre><pre name="code" class="cpp">#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;#define N 1005int h[N],a[N];int n,m;int cmp(int a,int b){    return  a>b;}int main(){    int i,j;    char c;    while(~scanf("%d%d",&n,&m))    {        getchar();        memset(h,0,sizeof(h));       int ans=0;       for(i=1;i<=n;i++)       {           for(j=1;j<=m;j++)           {               scanf("%c",&c);               if(c=='1')                 h[j]++;               else                 h[j]=0;           }           getchar();           for(j=1;j<=m;j++)             a[j]=h[j];           sort(a+1,a+m+1,cmp);  //强调一次，是一列一列的移动           for(j=1;j<=m;j++)            ans=max(ans,a[j]*j);       }        printf("%d\n",ans);    }    return 0;}