void dfs(int node) {
REP(i, 0, 6) {
if (m[node][i] != 0) {
//枚举该点所连路径
m[node][i]--;
m[i][node]--;
//递归
dfs(i);
//递归后存边
cnt++;
c[cnt].first = node;
c[cnt].second = i;
}
}
}
for (int i = cnt; i > 0; i--) {
//答案为倒序的
}