Masha wants to open her own bakery and bake muffins in one of the n cities numbered from 1 to n. There are m bidirectional roads, each of whose connects some pair of cities.
To bake muffins in her bakery, Masha needs to establish flour supply from some storage. There are onlyk storages, located in different cities numbered a1, a2, ..., ak.
Unforunately the law of the country Masha lives in prohibits opening bakery in any of the cities which has storage located in it. She can open it only in one of another n - k cities, and, of course, flour delivery should be paid — for every kilometer of path between storage and bakery Masha should pay 1 ruble.
Formally, Masha will pay x roubles, if she will open the bakery in some city b (ai ≠ b for every 1 ≤ i ≤ k) and choose a storage in some city s (s = aj for some 1 ≤ j ≤ k) and b and s are connected by some path of roads of summary length x (if there are more than one path, Masha is able to choose which of them should be used).
Masha is very thrifty and rational. She is interested in a city, where she can open her bakery (and choose one of k storages and one of the paths between city with bakery and city with storage) and pay minimum possible amount of rubles for flour delivery. Please help Masha find this amount.
The first line of the input contains three integers n, m and k (1 ≤ n, m ≤ 105, 0 ≤ k ≤ n) — the number of cities in country Masha lives in, the number of roads between them and the number of flour storages respectively.
Then m lines follow. Each of them contains three integers u, v and l (1 ≤ u, v ≤ n, 1 ≤ l ≤ 109, u ≠ v) meaning that there is a road between cities u and v of length of l kilometers .
If k > 0, then the last line of the input contains k distinct integers a1, a2, ..., ak (1 ≤ ai ≤ n) — the number of cities having flour storage located in. If k = 0 then this line is not presented in the input.
Print the minimum possible amount of rubles Masha should pay for flour delivery in the only line.
If the bakery can not be opened (while satisfying conditions) in any of the n cities, print - 1 in the only line.
5 4 2 1 2 5 1 2 3 2 3 4 1 4 10 1 5
3
3 1 1 1 2 3 3
-1
Image illustrates the first sample case. Cities with storage located in and the road representing the answer are darkened.
题目链接：CF 707B
多起点多终点求其中与终点相连的所有最短路中的最短长度，感觉跟优先队列有关，然后设一开始的所有面粉点即不能达到的d为0，其他都设为INF，把所有的面粉点压入队列进行BFS，最后选出1~n点中不为0的最短距离……一开始没考虑重边而且面粉店的d设的不对，WA两发……看了下其他人写的好像直接暴力比较（因为最短肯定是直接相连）而且速度跟BFS一样，我的做法比较麻烦……
代码：
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<bitset>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=100010;
struct info
{
int to;
int pre;
LL dx;
};
struct A
{
int cur;
LL dx;
bool operator<(const A &t)const
{
return dx>t.dx;
}
};
A pos[N];
info E[N<<1];
int head[N<<1],ne;
LL d[N];
priority_queue<A>Q;
void init()
{
CLR(head,-1);
ne=0;
CLR(d,INF);
while (!Q.empty())
Q.pop();
}
void add(int s,int t,LL dx)
{
E[ne].to=t;
E[ne].dx=dx;
E[ne].pre=head[s];
head[s]=ne++;
}
void bfs(int k)
{
int i;
for (i=0; i<k; ++i)
{
Q.push(pos[i]);
d[pos[i].cur]=0;
}
while (!Q.empty())
{
A now=Q.top();
Q.pop();
for (i=head[now.cur]; ~i; i=E[i].pre)
{
A v=now;
v.cur=E[i].to;
v.dx+=E[i].dx;
if(d[v.cur]>v.dx)
{
d[v.cur]=v.dx;
Q.push(v);
}
}
}
}
int main(void)
{
int n,m,k,i,j,a,b,temp;
LL dx;
while (~scanf("%d%d%d",&n,&m,&k))
{
init();
for (i=0; i<m; ++i)
{
scanf("%d%d%I64d",&a,&b,&dx);
add(a,b,dx);
add(b,a,dx);
}
for (i=0; i<k; ++i)
{
scanf("%d",&temp);
pos[i].cur=temp;
pos[i].dx=0;
}
bfs(k);
LL ans=d[0];
for (i=1; i<=n; ++i)
{
if(d[i]&&d[i]<ans)
ans=d[i];
}
printf("%I64d
",ans!=d[0]?ans:-1LL);
}
return 0;
}