Matrix Swapping II

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1543    Accepted Submission(s): 1036

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.

题解：

也是求最大矩阵的，只不过可以相互交换任意两列。

访问每一行时，求出每个点的高度，然后排序，以该点为高时的宽度就很容易看出来，面积取最大的就可以了。

ans=max(ans,(k-j+1)*s[j]);由于可以交换列，这里随着往右走，宽度逐渐减小，找最大值；

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define SD(x) scanf("%lf",&x)
#define P_ printf(" ")
typedef long long LL;
const int MAXN=1010;
char mp[MAXN][MAXN];
int dp[MAXN][MAXN];
int s[MAXN];
int main(){
    int N,M;
    while(~scanf("%d%d",&N,&M)){
        mem(dp,0);
        int ans=0;
        for(int i=1;i<=N;i++){
            scanf("%s",mp[i]);
            for(int j=M;j>=1;j--){
                mp[i][j]=mp[i][j-1];
            }
        }
        for(int i=1;i<=N;i++){
            int k=0;
            for(int j=1;j<=M;j++){
                if(mp[i][j]==‘1‘){
                    dp[i][j]=dp[i-1][j]+1;
                    s[++k]=dp[i][j];
                }
            }
            sort(s+1,s+k+1);
            for(int j=1;j<=k;j++){
                ans=max(ans,(k-j+1)*s[j]);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}