【LOJ #3087】【GXOI / GZOI2019】—旅行者（最短路）

传送门

想了一个跑一次$dij$的做法
然后华丽丽的被叉掉了
这样还有$80pts$

正解直接正反跑2次$dij$染色就可以了

#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ob==ib)?EOF:*ib++;
}
inline int read(){
    char ch=gc();
    int res=0,f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
#define ll long long
#define re register
#define pii pair<int,ll>
#define fi first
#define se second
#define pb push_back
#define cs const
#define bg begin
#define poly vector<int>  
inline void chemx(int &a,int b){a<b?a=b:0;}
inline void chemn(ll &a,ll b){a>b?a=b:0;}
cs int N=100005;
int n,m,k;
vector<pii> e[N],pe[N];
ll dis1[N],dis2[N];
int vis[N],tag[N],ist[N],pre1[N],pre2[N];
priority_queue<pii,vector<pii>,greater<pii> >q;
inline void dijkstra(ll *dis,int *pre){
	memset(dis,127/4,sizeof(ll)*(n+1));
	memset(vis,0,sizeof(vis));
	for(int i=1;i<=k;i++)
		dis[tag[i]]=0,q.push(pii(0,tag[i])),pre[tag[i]]=tag[i];
	while(!q.empty()){
		int u=q.top().se;q.pop();
		if(vis[u])continue;
		vis[u]=1;
		for(pii &x:e[u]){
			int v=x.fi;
			if(dis[u]+x.se<dis[v]){
				dis[v]=dis[u]+x.se;
				pre[v]=pre[u];
				q.push(pii(dis[v],v));
			}
		}
	}
}
inline void solve(){
	n=read(),m=read(),k=read();
	for(int i=1;i<=m;i++){
		int u=read(),v=read(),w=read();
		pe[u].pb(pii(v,w)),e[v].pb(pii(u,w));
	}
	for(int i=1;i<=k;i++)tag[i]=read(),ist[tag[i]]=1;
	dijkstra(dis1,pre1);
	for(int i=1;i<=n;i++)e[i]=pe[i];
	dijkstra(dis2,pre2);
	ll res=1e18;
	for(int u=1;u<=n;u++)
	for(pii &x:e[u]){
		if(pre1[x.fi]&&pre2[u]&&pre1[x.fi]!=pre2[u])chemn(res,dis1[x.fi]+dis2[u]+x.se);
	}
	cout<<res<<'
';
	for(int i=1;i<=n;i++)ist[i]=0,e[i].clear(),pe[i].clear(),pre1[i]=pre2[i]=0;
}
int main(){
	#ifdef Stargazer
	freopen("lx.cpp","r",stdin);
	#endif
	int T=read();
	while(T--)solve();
}