Codeforces 1385E 拓扑排序(判环)

拓扑玩少了想不到

题意

  给了n个点,m个边,其中有些边是有向边,有些是无向边。要求你给无向边规定方向,使得该图无换环。如果不能则输出-1。否则输出这m个边的顶点u,v,要求u指向v。

思路

  用拓扑排序判断环。如果出现的点的次数不等于n,则是无解。

  学大佬个小技巧:给每个点定义一个nub记录出现的前后顺序,最后输出无向边指向的时候,一定是nub小的指向nub大的(即先出现的指向后出现的)

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector> 
// #include <bits/stdc++.h>
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define sp ' '
#define endl '\n'
#define inf  0x3f3f3f3f;
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
#define P pair<int, int>
#define fi first
#define se second
#define pb(x) push_back(x)
#define ppb() pop_back()
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define repd(i,a,b) for(int i=a;i>=b;i--)
#define sca3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define sca2(a,b) scanf("%d %d",&(a),&(b))
#define sca(a) scanf("%d",&(a));
#define sca3ll(a,b,c) scanf("%lld %lld %lld",&(a),&(b),&(c))
#define sca2ll(a,b) scanf("%lld %lld",&(a),&(b))
#define scall(a) scanf("%lld",&(a));


using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a, ll b, ll mod){ll sum = 1;while (b) {if (b & 1) {sum = (sum * a) % mod;b--;}b /= 2;a = a * a % mod;}return sum;}

const double Pi = acos(-1.0);
const double epsilon = Pi/180.0;
const int maxn = 3e5+10;
int n,m;
int de[maxn],nub[maxn];
vector<int>e[maxn];
queue<pair<int,int> >ans;
int flag = 0;

void Init()
{
    cin>>n>>m;
    rep(i,1,n) {
        de[i] = 0;
        nub[i] = 0;
        e[i].clear();
    }
    rep(i,1,m){
        int u,v,x;
        cin>>x>>u>>v;
        if(x == 1){
            e[u].pb(v);
            de[v]++;
        }
        ans.push(mp(u,v));
    }
}
void Topsort()
{
    flag = 1;
    queue<int>que;
    rep(i,1,n){
        if(!de[i]){
            que.push(i);
        }
    }
    int cnt = 0;
    while(que.size()){
        int x = que.front();que.pop();
        nub[x] = ++cnt;
        for(int i: e[x]){
            de[i]--;
            if(de[i] == 0){
                que.push(i);
            }
        } 
    }
    if(cnt!=n) //是no
        flag=0;
}
int main()
{
    //freopen("input.txt", "r", stdin);
    int _;
    scanf("%d",&_);
    while(_--)
    {
        Init();
        Topsort();
        if(flag==0){
            cout<<"NO"<<endl;
            while(ans.size()) ans.pop();
            continue;
        }
        cout<<"YES"<<endl;
        while(ans.size()){
            int u = ans.front().fi;
            int v = ans.front().se;
            ans.pop();
            if(nub[u] > nub[v]) swap(u,v);
            cout<<u<<sp<<v<<endl;
        }
    }
}

 

posted @ 2020-07-22 17:55  阿斯水生产线  阅读(204)  评论(0编辑  收藏  举报