https://vjudge.net/problem/Kattis-slowleak
题意：
从1到n，路径长度。有多个点可以加油，每次加完油后只能走一定的路程。500个点。
解析：
先做一遍Floyd，然后把加油站抽出来再做一遍Floyd即可。
代码：
/*
* Author : Jk_Chen
* Date : 2020-02-06-14.51.14
*/
#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define rep(i,a,b) for(int i=(int)(a);i<=(int)(b);i++)
#define per(i,a,b) for(int i=(int)(a);i>=(int)(b);i--)
#define mmm(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define pill pair<int, int>
#define fi first
#define se second
#define debug(x) cerr<<#x<<" = "<<x<<'\n'
const LL mod=1e9+7;
const int maxn=503;
const int inf=0x3f3f3f3f;
LL rd(){ LL ans=0; char last=' ',ch=getchar();
while(!(ch>='0' && ch<='9'))last=ch,ch=getchar();
while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();
if(last=='-')ans=-ans; return ans;
}
#define rd rd()
/*_________________________________________________________begin*/
int n,m,k;
LL V;
vector<int>pos;
LL dis[maxn][maxn];
LL INF;
void floyd(){
rep(K,1,n)
rep(i,1,n){
rep(j,1,n){
dis[j][i]=dis[i][j]=min(dis[i][j],dis[i][K]+dis[K][j]);
}
}
rep(i,0,pos.size()-1){
rep(j,0,pos.size()-1){
if(dis[pos[i]][pos[j]]>V){
dis[pos[i]][pos[j]]=INF;
}
}
}
rep(K,0,pos.size()-1)
rep(i,0,pos.size()-1){
rep(j,0,pos.size()-1){
dis[pos[j]][pos[i]]=dis[pos[i]][pos[j]]=min(dis[pos[i]][pos[j]],dis[pos[i]][pos[K]]+dis[pos[K]][pos[j]]);
}
}
}
int main(){
mmm(dis,0x3f);
INF=dis[0][0];
n=rd,m=rd,k=rd,V=rd;
rep(i,1,n)dis[i][i]=0;
rep(i,1,k){
int p=rd;
pos.pb(p);
}
pos.pb(1);
pos.pb(n);
rep(i,1,m){
int a=rd,b=rd;
LL v=rd;
dis[a][b]=dis[b][a]=v;
}
floyd();
if(dis[1][n]>=INF){
puts("stuck");
}
else{
printf("%lld\n",dis[1][n]);
}
}