hdu2830 Matrix Swapping II（DP）

来源：互联网发布：太平洋炒股软件下载编辑：程序博客网时间：2024/04/30 03:34

题意：给你一个矩阵，里面的数字只有0和1两种，其中，列可以任意移动。问如何移动可以使某个子矩阵中元素全部是1，求出这个最大子矩阵的面积。

思路：由于列可以任意移动，那么显然把高度大的都放到一堆里面是最优的，那么排个序就可以像HDU1506那样做了

#include<bits\stdc++.h>using namespace std;const int maxn = 2005;char mp[maxn][maxn];int h[maxn];int hh[maxn];bool cmp(int a,int b){return a>b;}int main(){     int n,m; while(scanf("%d%d",&n,&m)!=EOF) { int ans = 0;// getchar(); for(int i = 1;i<=n;i++) {// memset(h,0,sizeof(h)); for(int j = 1;j<=m;j++) { scanf(" %c",&mp[i][j]); if(mp[i][j]=='1') h[j]++; else h[j]=0; hh[j]=h[j]; } sort(hh+1,hh+1+m,cmp); for (int j = 1;j<=m;j++) ans = max(ans,hh[j]*j);// getchar(); } printf("%d\n",ans); }}

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.

0 0