HDU 2830 Matrix Swapping II （最大子矩阵）

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Matrix Swapping II

Time Limit: 9000/3000 MS (Java/Others)

Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 809 Accepted Submission(s): 547

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

View Code

  1 /*
  2   这道题跟1505,1506,2870又是类似的。求最大矩阵。
  3   不同的是，这里的每一列都是可以移动的，所以不必找左右边界的问题，对于
  4   每一层，统计这一层中每一列的高度，然后将这些高度按降序排序，就解决了
  5   列可以移动的问题，最后答案ans = max(ans,num[i]*i);
  6   
  7   328MS    5168K
  8   */
  9 #include <iostream>
 10 #include <cstdio>
 11 #include <cstring>
 12 #include <algorithm>
 13 #define SIZE 1005
 14 
 15 using namespace std;
 16 
 17 char ch[SIZE][SIZE];
 18 int height[SIZE][SIZE];
 19 int num[SIZE];
 20 int n,m;
 21 
 22 bool cmp(int a,int b)
 23 {
 24     return a > b;
 25 }
 26 
 27 int main()
 28 {
 29     while(~scanf("%d%d",&n,&m))
 30     {
 31         int ans = 0;
 32         for(int i=1; i<=n; i++)
 33         {
 34             scanf("%s",ch[i]+1);
 35         }
 36         for(int i=1; i<=n; i++)
 37         {
 38             for(int j=1; j<=m; j++)
 39             {
 40                 if(ch[i][j] == '1')
 41                     height[i][j] = height[i-1][j]+1;
 42                 else
 43                     height[i][j] = 0;
 44                 num[j] = height[i][j];
 45             }
 46             sort(num+1,num+1+m,cmp);
 47             for(int i=1; i<=m; i++)
 48                 ans = max(ans,num[i]*i);
 49         }
 50         printf("%d\n",ans);
 51     }
 52     return 0;
 53 }
 54 /*
 55   网上看到的空间优化，而且代码更简洁，值得学习，按照这个思路，再打了一遍。
 56   
 57   312MS    240K
 58   
 59 #include <iostream>
 60 #include <cstdio>
 61 #include <cstring>
 62 #include <algorithm>
 63 #define SIZE 1005
 64 
 65 using namespace std;
 66 
 67 char ch[SIZE];
 68 int height[SIZE];
 69 int num[SIZE];
 70 int n,m;
 71 
 72 bool cmp(int a,int b)
 73 {
 74     return a > b;
 75 }
 76 
 77 int main()
 78 {
 79     while(~scanf("%d%d",&n,&m))
 80     {
 81         memset(height,0,sizeof(height));
 82         int ans = 0;
 83         for(int i=1; i<=n; i++)
 84         {
 85             scanf("%s",ch+1);
 86             for(int j=1; j<=m; j++)
 87             {
 88                 if(ch[j] == '1')
 89                     height[j] ++;
 90                 else
 91                     height[j] = 0;
 92                 num[j] = height[j];
 93             }
 94             sort(num+1,num+1+m,cmp);
 95             for(int i=1; i<=m; i++)
 96                 ans = max(ans,num[i]*i);
 97         }
 98         printf("%d\n",ans);
 99     }
100     return 0;
101 }
102 */