HDu 2830 Matrix Swapping II（dp）

开发者柚溪技术 2022年11月18日

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Problem DescriptionGiven 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.
InputThere 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
OutputOutput one line for each test case, indicating the maximum possible goodness. Sample Input

Sample Output

4
2Note: 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,然后就能够求最优解了

</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;
}