一、内容

 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

二、思路

三、代码

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 105;
struct Node{ int c1, c2;} ans[N * 10];
int n, g[N][N], mat[N]; //代表某列匹配的是哪行 
bool vis[N];
bool dfs(int  u) {
	for (int v = 1; v <= n; v++) {
		if (vis[v] || !g[u][v]) continue; vis[v] = true;
		if (!mat[v] || dfs(mat[v])) {
			mat[v] = u; return true;
		}
	}
	return false;
} 
int main() {
	while (~scanf("%d", &n)) {
		memset(g, 0, sizeof(g));
		memset(mat, 0, sizeof(mat));
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) scanf("%d", &g[i][j]); 
		} 
		int cnt = 0, m = 0;
		for (int i = 1; i <= n; i++) {
			memset(vis, false, sizeof(vis));
			if (dfs(i)) cnt++;
		} 
		if (cnt != n) {printf("-1\n"); continue;}
		//交换2列 知道(c, c)的位置为1 
		for (int c = 1; c <= n; c++) {
			while (c != mat[c]) {
				ans[++m].c1 = c; 
				ans[m].c2 = mat[c]; //将（mat[c], c）的1 交换到(mat[c], mat[c])
				//那么行和列 匹配 也要交换
				swap(mat[c], mat[mat[c]]); 
			} 
		}
		printf("%d\n", m);
		for (int i = 1; i <= m; i++) printf("C %d %d\n", ans[i].c1, ans[i].c2);
	}
	return 0;
}