codeforces-1385E(拓扑排序)

Directing Edges

题目描述：

给定n个点m条边，其中一些边是有向的，一些边是无向的，现在你需要将这些无向的边确定方向，并判断是否可以生成一个有向无环图

思路：

显而易见如果给出的有向边没有形成环的话，剩下的无向边一定可以使他们不形成环，于是只需要将给定的有向边做一遍拓扑排序，判断是否已经成环，未成环的话就将所有无向边按拓扑序定向即可使图无环

代码：

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
const ll N = 1e6;
const double PI = acos(-1.0);
#define Test ll tesnum;tesnum = read();while(tesnum--)
ll read();
vector<int> g[200005];
vector<pair<int,int>> v;
int deg[N],pos[N];
int cnt;
int main()
{
    Test{
        cnt = 0;
        v.clear();
        int n,m;
        cin>>n>>m;
        for(int i = 1; i <= n; i++)g[i].clear();
        for(int i = 1; i <= n; i++)deg[i] = pos[i] = 0;
        for(int i = 1; i <= m; i++){
            int w,x,y;
            cin>>w>>x>>y;
            v.emplace_back(make_pair(x,y));
            if(w==1){
                g[x].emplace_back(y);
                deg[y]++;
            }
        }
        queue<int> q;
        for(int i = 1; i <= n; i++){
            if(deg[i]==0)
                q.push(i);
        }
        while(!q.empty())
        {
            int u = q.front();
            q.pop();
            pos[u] = ++cnt;
            for(int i = 0; i < g[u].size(); i++){
                deg[g[u][i]]--;
                if(deg[g[u][i]]==0){
                    q.push(g[u][i]);
                }
            }
        }
        if(cnt==n){
            cout<<"YES"<<endl;
            for(auto x:v){
                if(pos[x.first]>pos[x.second]){
                    cout<<x.second<<" "<<x.first<<endl;
                }else
                    cout<<x.first<<" "<<x.second<<endl;
            }
        }else{
            cout<<"NO"<<endl;
        }

    };
    return "BT7274", NULL;
}

inline ll read() {
    ll hcy = 0, dia = 1;char boluo = getchar();
    while (!isdigit(boluo)) {if (boluo == '-')dia = -1;boluo = getchar();}
    while (isdigit(boluo)) {hcy = hcy * 10 + boluo - '0';boluo = getchar();}
    return hcy * dia;
}