题目链接

  题意：N个点，M条边，其中“1”边是有向边，“0”边是无向边，我们要给无向边定向，使得这个图不会构成环。

  构成环会让我们想到关于拓扑排序，因为如果成环了，就会使得有点不能出拓扑了，所以当我们跑拓扑序的时候，我们可以给无向边的指向定为从拓扑小的指向拓扑序大的。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 2e5 + 7;
int N, M, head[maxN], cnt;
struct Eddge
{
    int nex, to;
    Eddge(int a=-1, int b=0):nex(a), to(b) {}
} edge[maxN << 1];
inline void addEddge(int u, int v)
{
    edge[cnt] = Eddge(head[u], v);
    head[u] = cnt++;
}
vector<pair<int, int>> vt;
int que[maxN], top, tail, dfn[maxN], tot, du[maxN];
void Tp_sort()
{
    top = tail = 0;
    for(int i=1; i<=N; i++) if(!du[i]) que[tail++] = i;
    while(top < tail)
    {
        int u = que[top++];
        dfn[u] = ++tot;
        for(int i=head[u], v; ~i; i=edge[i].nex)
        {
            v = edge[i].to;
            if(!(--du[v])) que[tail++] = v;
        }
    }
}
inline void init()
{
    cnt = tot = 0; vt.clear();
    for(int i=1; i<=N; i++) { head[i] = -1; dfn[i] = du[i] = 0; }
}
int main()
{
    int T; scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &N, &M);
        init();
        for(int i=1, op, u, v; i<=M; i++)
        {
            scanf("%d%d%d", &op, &u, &v);
            if(!op) vt.push_back(make_pair(u, v));
            else { addEddge(u, v); du[v]++; }
        }
        Tp_sort();
        bool ok = true;
        for(int i=1; i<=N; i++) if(!dfn[i]) { ok = false; break; }
        if(ok)
        {
            printf("YES\n");
            int len = (int)vt.size();
            for(int i=0, u, v; i<len; i++)
            {
                u = vt[i].first; v = vt[i].second;
                if(dfn[u] > dfn[v]) printf("%d %d\n", v, u);
                else printf("%d %d\n", u, v);
            }
            for(int u=1; u<=N; u++)
            {
                for(int i=head[u], v; ~i; i=edge[i].nex)
                {
                    v = edge[i].to;
                    printf("%d %d\n", u, v);
                }
            }
        }
        else printf("NO\n");
    }
    return 0;
}