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
#include<bits/stdc++.h>
using namespace std;
const int M=110;
int n,m[M],f[M],vis[M];
int g[M][M],ans[M][2];
bool dfs(int x){
int i;
for(i=1;i<=n;i++)
if(g[x][i]&&vis[i]==0){
vis[i]=1;
if(m[i]==0||dfs(m[i])){
m[i]=x;
return 1;
}
}
return 0;
}
int main(){
int l,s,i,j,k;
while(scanf("%d",&n)!=EOF){
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
scanf("%d",&g[i][j]);
memset(m,0,sizeof(m));
s=0;
for(i=1;i<=n;i++){
memset(vis,0,sizeof(vis));
if(dfs(i))s++;
}
if(s<n)printf("-1\n");
else{
l=0;
for(i=1;i<=n;i++)
f[i]=i;
for(i=1;i<=n;i++)
if(m[i]!=f[i]){
for(j=i+1;j<=n;j++)
if(m[i]==f[j])
break;
l++;
ans[l][0]=i;
ans[l][1]=j;
f[j]=f[i];
}
printf("%d\n",l);
for(i=1;i<=l;i++)
printf("R %d %d\n",ans[i][0],ans[i][1]);
}
}
}
