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

Posted 2022-12-01 TURNINING

传送门

题意：一个图，给你n个顶点，m条边，m条边中有些是有向的，有些是无向的。现在要求你把无向的边全部变成有向的边，能否使图最终无环？

思路：先只看有向边，若原图已经有环那么肯定无解，若无环我们一定能通过拓扑序构造一个图使图无环。

证明：先跑一个拓扑排序，对于一个无向的边，我们只要把边：从拓扑序小的指向拓扑序大的 就能保正不产生环(拓扑序的定义！)

#include<bits/stdc++.h>
using namespace std;

#define lsn (u << 1)
#define rsn (u << 1 | 1)
#define mid (l + r >> 1)

typedef long long ll;
typedef unsigned long long ull;

typedef pair<int, int> P;

const int MAXN = 2e5 + 10;
const int MAX_LEN = 100000 + 10;
const int MAX_LOG_V = 22;
const int INF = 1e6;
const int mod = 1e9 + 7;
const double eps = 1e-7;
const ull B = 100000007;


int n, m, no[MAXN], d[MAXN];
vector<int> g[MAXN];
P f[MAXN];


void solve() 
	scanf("%d %d", &n, &m);
	for(int i = 1; i <= n; i++)  d[i] = 0; g[i].clear(); 
	for(int i = 1; i <= m; i++) 
		int op, u, v; scanf("%d %d %d", &op, &u, &v);
		f[i].first = u; f[i].second = v;
		if(op) 
			g[u].emplace_back(v);
			d[v]++;
		
	
	queue<int> q;
	for(int i = 1; i <= n; i++) 
		if(!d[i]) q.emplace(i); 
	
	int tot = 0;
	while(!q.empty()) 
		int u = q.front(); q.pop();
		no[u] = ++tot;
		for(int i = 0; i < g[u].size(); i++) 
			int v = g[u][i];
			d[v]--;
			if(!d[v]) q.emplace(v);
		
	
	if(tot != n) puts("NO");
	else 
		puts("YES");
		//printf("YES\\n");
		for(int i = 1; i <= m; i++) 
			if(no[f[i].first] < no[f[i].second]) printf("%d %d\\n", f[i].first, f[i].second);
			else printf("%d %d\\n", f[i].second, f[i].first);
			
	


int main() 
	int t = 1; scanf("%d", &t);
	while(t--)  
		solve();