D. Exploration plan

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output

The competitors of Bubble Cup X gathered after the competition and discussed what is the best way to get to know the host country and its cities.

After exploring the map of Serbia for a while, the competitors came up with the following facts: the country has V cities which are indexed with numbers from 1 to V, and there are E bi-directional roads that connect the cites. Each road has a weight (the time needed to cross that road). There are N teams at the Bubble Cup and the competitors came up with the following plan: each of the N teams will start their journey in one of the V cities, and some of the teams share the starting position.

They want to find the shortest time T, such that every team can move in these T minutes, and the number of different cities they end up in is at least K (because they will only get to know the cities they end up in). A team doesn’t have to be on the move all the time, if they like it in a particular city, they can stay there and wait for the time to pass.

Please help the competitors to determine the shortest time T so it’s possible for them to end up in at least K different cities or print -1 if that is impossible no matter how they move.

Note that there can exist multiple roads between some cities.

Input
The first line contains four integers: V, E, N and K (1 ≤  V  ≤  600,  1  ≤  E  ≤  20000,  1  ≤  N  ≤  min(V, 200),  1  ≤  K  ≤  N), number of cities, number of roads, number of teams and the smallest number of different cities they need to end up in, respectively.

The second line contains N integers, the cities where the teams start their journey.

Next E lines contain information about the roads in following format: Ai Bi Ti (1 ≤ Ai, Bi ≤ V,  1 ≤ Ti ≤ 10000), which means that there is a road connecting cities Ai and Bi, and you need Ti minutes to cross that road.

Output
Output a single integer that represents the minimal time the teams can move for, such that they end up in at least K different cities or output -1 if there is no solution.

If the solution exists, result will be no greater than 1731311.

Example
inputCopy
6 7 5 4
5 5 2 2 5
1 3 3
1 5 2
1 6 5
2 5 4
2 6 7
3 4 11
3 5 3

outputCopy
3

Note
Three teams start from city 5, and two teams start from city 2. If they agree to move for 3 minutes, one possible situation would be the following: Two teams in city 2, one team in city 5, one team in city 3 , and one team in city 1. And we see that there are four different cities the teams end their journey at.

题目大意：有很多支队伍，从一些城市出发，问最后到达不同城市数量为k的最小时间，或者不能到达。

二分时间，我们对于所有城市来说，如果这个队伍所在城市能够到达一些城市并且时间小于mid，就可以连边，最后每个城市向汇点连边，流量为1.每次二分check的时候判断最大流即可。

AC代码：

#pragma GCC optimize(2)
#include<bits/stdc++.h>
//#define int long long
using namespace std;
const int inf=0x3f3f3f3f;
const int N=810,M=1e6+10;
int n,m,v,k,h[N],s,t,st[N],g[N][N];
int head[N],nex[M],to[M],w[M],tot;
inline void ade(int a,int b,int c){
	to[++tot]=b; nex[tot]=head[a]; w[tot]=c; head[a]=tot;
}
inline void add(int a,int b,int c){
	ade(a,b,c);	ade(b,a,0);
}
void floyd(){
	for(int k=1;k<=n;k++)
		for(int i=1;i<=n;i++)
			for(int j=1;j<=n;j++)
				g[i][j]=min(g[i][j],g[i][k]+g[k][j]);
}
inline int bfs(){
	queue<int> q;	q.push(s); 	memset(h,0,sizeof h); h[s]=1;
	while(q.size()){
		int u=q.front();	q.pop();
		for(int i=head[u];i;i=nex[i]){
			if(w[i]&&!h[to[i]]){
				h[to[i]]=h[u]+1;	q.push(to[i]);
			}
		}
	}
	return h[t];
}
int dfs(int x,int f){
	if(x==t)	return f; int fl=0;
	for(int i=head[x];i&&f;i=nex[i]){
		if(w[i]&&h[to[i]]==h[x]+1){
			int mi=dfs(to[i],min(w[i],f));
			w[i]-=mi; w[i^1]+=mi; fl+=mi; f-=mi;
		}
	}
	if(!fl)	h[x]=-1;
	return fl;
}
int dinic(){
	int res=0;
	while(bfs())	res+=dfs(s,inf);
	return res;
}
int check(int mid){
	tot=1;	memset(head,0,sizeof head);
	for(int i=1;i<=v;i++){
		add(s,i,1);
		for(int j=1;j<=n;j++){
			if(g[st[i]][j]<=mid)	add(i,j+v,1);
		}
	}
	for(int i=1;i<=n;i++)	add(i+v,t,1);
	return dinic()>=k;
}
signed main(){
	cin>>n>>m>>v>>k; t=n+v+1;
	for(int i=1;i<=n;i++)	for(int j=1;j<=n;j++)	g[i][j]=(i==j?0:inf);
	for(int i=1;i<=v;i++)	scanf("%d",&st[i]);
	for(int i=1;i<=m;i++){
		int a,b,c;	scanf("%d %d %d",&a,&b,&c);
		g[a][b]=g[b][a]=min(g[a][b],c);
	}
	floyd();
	int l=0,r=1731313;
	while(l<r){
		int mid=l+r>>1;
		if(check(mid))	r=mid;
		else	l=mid+1;
	}
	printf("%d\n",r>1731311?-1:r);
	return 0;
}