标签:ret int lang pos 生成 lse bit har 排序
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
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
#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;
}
标签:ret int lang pos 生成 lse bit har 排序
原文地址:https://www.cnblogs.com/cloudplankroader/p/13341693.html