Hdu 2830 Matrix Swapping II【思维】

Matrix Swapping II

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

Total Submission(s): 1861    Accepted Submission(s): 1239


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.

题目大意：

我们可以任意交换两列，我们要找到一个最大面积的矩阵，使得其中所有元素都是1.

思路：

我们处理出以第i行为基础水平线的向下延展的柱形高度(以样例1为基础)sum【i】【j】：

2 0 1 3

1 0 0 2

0 0 0 1

然后我们以每一行为枚举量，去sort这一行的sum【i】【j】（从大到小）；

那么ans=max（ans，j*sum【i】【j】）；

Ac代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
char a[1005][1005];
int sum[1005][1005];
int vis[1005];
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
memset(sum,0,sizeof(sum));
for(int i=0;i<n;i++)
{
scanf("%s",a[i]);
}
for(int i=0;i<m;i++)
{
for(int j=n-1;j>=0;j--)
{
if(a[j][i]=='1')
{
sum[j][i]=sum[j+1][i]+1;
}
}
}
int output=0;
for(int i=0;i<n;i++)
{
memset(vis,0,sizeof(vis));
for(int j=0;j<m;j++)
{
vis[sum[i][j]]++;
}
int tmp=0;
for(int j=n;j>=1;j--)
{
tmp+=vis[j];
output=max(output,tmp*j);
}
}
printf("%d\n",output);
}
}