杭电 2830 数学题

         题目：

Matrix Swapping II

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 512    Accepted Submission(s): 346

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.

 

ac代码：

/*用a[i]代表矩形的宽，用t代表矩形的长，则两者的积就是矩形的面积。用sort对b数组排序的过程其实就是交换列的过程。*/#include <iostream>#include <algorithm>#include <string.h>#include <string>using namespace std;#define M 1010int a[M],b[M];char ch[M];void init(){  memset(a,0,sizeof(a));  memset(b,0,sizeof(b));  memset(ch,0,sizeof(ch));}int main(){  int n,m;  while(~scanf("%d%d",&n,&m)){    init();int ans=0;for(int i=0;i<n;++i){  scanf("%s",ch);  for(int j=0;j<m;++j){    if(ch[j]=='1')a[j]++;elsea[j]=0;//如果列上的1不连续的话，则宽为0  }  for(int j=0;j<m;++j)  b[j]=a[j];  sort(b,b+m);  int t=-1,sum=0;  for(int j=0;j<m;++j){  if(t!=b[j]){    t=b[j];sum=t*(m-j);if(sum>ans)ans=sum;  }  }}printf("%d\n",ans);  }  return 0;}