一道类似缩点的好题，先按道路缩点 然后将缩点以后的图按照航线做DAG

在DAG上先跑topsort 在每一个团内部跑dijkstra，同时更新top点 很有意思的一道题目

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 3e5+10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;int n,m1,m2,st;int e[N],ne[N],w[N],h[N],idx;
void add(int a,int b,int c){e[idx] = b,ne[idx] = h[a],w[idx] = c,h[a] = idx++;
}int din[N];
int id[N];
vector<int>block[N];
bool df[N];
int cnt;
int dist[N];
bool vis[N];void dfs(int u,int ids){df[u] = true;id[u] = ids;block[ids].push_back(u);for(int i=h[u];~i;i=ne[i]){int j = e[i];if(!df[j])dfs(j,ids);} 
}
queue<int>q;
void dijkstra(int ids)
{priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>>heap;for(auto &t:block[ids])heap.push({dist[t],t});while(heap.size()){auto t = heap.top();heap.pop();int ver = t.second,distance = t.first;if(vis[ver])continue;vis[ver] = true;for(int i=h[ver];~i;i=ne[i]){int j = e[i];//cout<<ver<<" "<<" "<<j<<"\n";if(dist[j]>distance+w[i]){dist[j] = dist[ver]+w[i];if(id[j]==id[ver]){heap.push({dist[j],j});}}if(id[j]!=id[ver]&&(--din[id[j]]==0))q.push(id[j]);}}}void topsort()
{for(int i=1;i<=cnt;++i)if(!din[i]){q.push(i);}memset(dist,0x3f,sizeof dist);memset(vis,0,sizeof vis); dist[st] = 0;while(q.size()){auto t = q.front();q.pop();//cout<<t<<"??\n";dijkstra(t);}}void solve()
{cin>>n>>m1>>m2>>st;memset(h,-1,sizeof h);for(int i=1;i<=m1;i++){int a,b,c;cin>>a>>b>>c;add(a,b,c),add(b,a,c);}//缩点for(int i=1;i<=n;i++)if(!df[i])dfs(i,++cnt);for(int i=1;i<=m2;i++){int a,b,c;cin>>a>>b>>c;add(a,b,c);din[id[b]]++;}topsort();for(int i=1;i<=n;i++)if(dist[i]>0x3f3f3f3f/2)cout<<"NO PATH\n";else cout<<dist[i]<<"\n";}signed main()
{ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);int _;//cin>>_;_ = 1;while(_--)solve();return 0;
}