Codeforces Round #656 (Div. 3) E. Directing Edges

给定一张图，图中给出一些有向边，一些无向边。

要求给所有无向边赋予方向后能够使整张图无环。

已知给定的图中无自环，无重边

解题思路：

若给定的有向边已经成环，那么必然输出NO，否则都可以构造出解。

考虑原图的拓扑排序，根据两点的进队时间，对于某一无向边，只要使边上两点进队时间也满足拓扑序，就可以使得图依旧无环。

#pragma warning(disable:4996)

#include<iostream>
#include<algorithm>
#include<bitset>
#include<tuple>
#include<unordered_map>
#include<fstream>
#include<iomanip>
#include<string>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<set>
#include<list>
#include<queue>
#include<stack>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<cstdlib>
#define INF 0x3f3f3f3f
#define inf 0x7FFFFFFF
#define MOD 998244353
#define moD 1000000003
#define pii pair<ll,int>
#define eps 1e-8
#define equals(a,b) (fabs(a-b)<eps)
#define bug puts("bug")
#define re  register
#define fi first
#define se second
#define pb push_back
const int maxn = 1e6 + 5;
const double Inf = 10000.0;
const double PI = acos(-1.0);
typedef  long long ll;
typedef unsigned long long ull;
using namespace std;

vector<int> vec[maxn];
int InDeg[maxn];
queue<int> q;
int num;
int n;
int t[maxn];

bool topsort() {
    while (!q.empty())  q.pop();
    num = 0;
    for (int i = 1; i <= n; i++) if (!InDeg[i]) q.push(i);
    int tt = 0;
    while (!q.empty()) {
        int now = q.front();
        q.pop();
        num++;
        t[now] = tt++;
        for (int i = 0; i < vec[now].size(); i++) {
            if (--InDeg[vec[now][i]] == 0) q.push(vec[now][i]);    //入度为零，进入队列

        }
    }
    return n == num;
}


void solve() {
    int tmp, x, y;
    memset(InDeg, 0, sizeof InDeg);
    memset(vec, 0, sizeof vec);
    memset(t, 0, sizeof t);
    int m;
    scanf("%d%d", &n, &m);
    vector<pii> ans;
    vector<pii> res;
    for (int i = 0; i < m; i++) {
        scanf("%d%d%d", &tmp, &x, &y);
        if (tmp == 1)  vec[x].push_back(y), InDeg[y]++, res.emplace_back(x, y);
        else ans.emplace_back(x, y);
    }
    if (!topsort()) {
        puts("NO");
        return;
    }
    for (int i = 0; i < ans.size(); i++) {
        if (t[ans[i].first] > t[ans[i].se]) res.emplace_back(ans[i].se, ans[i].fi);
        else if (t[ans[i].first] <= t[ans[i].se])res.emplace_back(ans[i].fi, ans[i].se);
    }
    puts("YES");
    for (int i = 0; i < res.size(); i++) printf("%d %d
", res[i].fi, res[i].se);
}



int main() {
    int T;
    scanf("%d", &T);
    while (T--) solve();
}