//做這道題很郁悶, 一開始以為計算行列式值為1.怎么做都不對. 后來發(fā)現(xiàn)只需要斜線都為1即可.
//就會想到二分圖的完全匹配
#include <iostream>
#include <queue>
#define MAXN 105
using namespace std;
typedef struct {
int u, v;
}change;
int n;
int re[MAXN][MAXN];
bool chk[MAXN];
int match[MAXN];
int find (int x) {
int i;
for (i = 1;i <= n;++ i) {
if (re[ x ][ i ] && !chk[ i ]) {
//若x與i相連,且i不在增廣路上
chk[ i ] = true;
if (!match[ i ] || find(match[ i ])) {
//若i還沒有匹配,或從i的匹配項有增廣路
match[ i ] = x;
return 1;
}
}
}
return 0;
}
void init() {
memset(match , 0 , sizeof(match));
}
bool res() {
int i, j, m, k, t;
for(i = 1; i <= n; ++ i) {
for(j = 1; j <= n; ++ j) {
scanf("%d", &re[ i ][ j ]);
}
}
init();
for (i = 1;i <= n;++ i) {
memset(chk , 0 , sizeof(chk));
if (!find(i)) return false;
}
queue<change> Q;
change p;
for (j = 1; j <= n; ++ j) {
m = j;
for (k = j; k <= n; ++ k) if (match[ k ] <= match[ m ]) m = k;
if (m != j) {
p.u = m; p.v = j;
Q.push(p);
t = match[ m ];
match[ m ] = match[ j ];
match[ j ] = t;
}
}
printf("%d\n", Q.size());
while(!Q.empty()) {
p = Q.front();
Q.pop();
printf("C %d %d\n", p.u, p.v);
}
return true;
}
int main() {
while (~scanf("%d", &n)) {
if(!res()) puts("-1");
}
return 0;
}