Swap——hdu 2819

Swap

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2152    Accepted Submission(s): 764
Special Judge

Problem Description

Given an N*N matrix with each entry equal to 0 or 1. You can swap any two rows or any two columns. Can you find a way to make all the diagonal entries equal to 1?

Input

There are several test cases in the input. The first line of each test case is an integer N (1 <= N <= 100). Then N lines follow, each contains N numbers (0 or 1), separating by space, indicating the N*N matrix.

Output

For each test case, the first line contain the number of swaps M. Then M lines follow, whose format is “R a b” or “C a b”, indicating swapping the row a and row b, or swapping the column a and column b. (1 <= a, b <= N). Any correct answer will be accepted, but M should be more than 1000.

If it is impossible to make all the diagonal entries equal to 1, output only one one containing “-1”.

Sample Input

2

0 1

1 0

2

1 0

1 0

Sample Output

1
R 1 2
-1

Source

2009 Multi-University Training Contest 1 - Host by TJU

 题意：如果可以交换行列，问主对角线能不能全为1

分析：要想主对角线全为1很明显要有N个行列不想同的点就行了，可以用二分图匹配计算出来多能有几个。如果小与N就不能。输出要是对的就行，不必和答案一样

 #include<iostream>
 #include<cstdio>
 #include<cstring>

 using namespace std;

 #define N 1100

 int n, vis[N], used[N], a[N], b[N];
 int maps[N][N];

 int found(int u)
 {
     for(int i = ; i <= n; i++)
     {
         if(!vis[i] && maps[u][i])
         {
             vis[i] = ;
             if(!used[i] || found(used[i]))
             {
                 used[i] = u;
                 return true;
             }
         }
     }
     return false;
 }
 int main()
 {
     int w;

     while(scanf("%d", &n) != EOF)
     {
         memset(vis, , sizeof(vis));
         memset(used, , sizeof(used));
         memset(maps, , sizeof(used));
         memset(a, , sizeof(a));
         memset(b, , sizeof(b));

         int ans = ;

         for(int i = ; i <= n; i++)
             for(int j = ; j <= n; j++)
                 scanf("%d", &maps[i][j]);

         for(int i = ; i <= n; i++)
         {
             memset(vis, , sizeof(vis));
             if(found(i))
                 ans++;
         }
         if(ans < n)
         {
             printf("-1\n");
             continue;
         }

         w = ;
         for(int i = ; i <= n; i++)
         {
             while(used[i] != i)
             {
                 a[w] = i;
                 b[w] = used[i];
                 swap(used[a[w]], used[b[w]]);  // 如果该行匹配不是自身，就交换匹配。
                 w++;
             }
         }
         printf("%d\n", w);
         for(int i = ; i < w; i++)
             printf("C %d %d\n", a[i], b[i]);

     }
     return ;
 }