hdu Swap

Swap

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 3   Accepted Submission(s) : 2

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

题目意思是说通过交换行或者列来实现对角线(左上角到右下角)上都是1，首先，如果某行全是0或者某列全是0必然不满足情况输出-1，还有就是对角线上的N个1，它们各自在不同的行中出现至少一次才可以，比如样例2中，虽然有两个1，但是它们总是处在同一列，仍然不满足要求，这时候问题的矛盾也比较明显了，很明显不能一个行对应两个列，或者说，每一行都应该有至少一个与自己对应的列才能满足条件，那么将行作为X集合，列作为Y集合，如果map[i][j]==1，那么Xi->Yj连边，求最大匹配，这样的话没有任何一个行被两个列匹配，也就满足了我们的要求，如果最大匹配==N，那么必然有解，否则必然误解，至于结果的输出，保存mx[]和my[]，表示与X匹配的点和与Y匹配的点，这样对于i（i∈[1,N]），如果mx[i]!=my[i]交换行i,my[i]就好了。