codeforces 707 B. Bakery (贪心)

最新推荐文章于 2021-03-14 09:47:01 发布

原创最新推荐文章于 2021-03-14 09:47:01 发布 · 930 阅读

0 ·

CC 4.0 BY-SA版权

codeforces 同时被 2 个专栏收录

36 篇文章

订阅专栏

贪心

35 篇文章

订阅专栏

本文介绍了一个关于面包店选址的问题，目标是在避开已有的面粉储存城市的前提下，找到距离面粉店最近的城市开设新店，通过贪心算法结合路径长度排序实现解决方案。

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 only k 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 citys (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.

Input

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.

Output

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.

Examples

input
5 4 2
1 2 5
1 2 3
2 3 4
1 4 10
1 5

output
3

input
3 1 1
1 2 3
3

output
-1

Note

$\text{[math]}$

Image illustrates the first sample case. Cities with storage located in and the road representing the answer are darkened.

题目大意：有n个城市，有个人想开面包，但是其中有k个是面粉店，某人想在n-k个城市中找一个离面粉店最近的城市,给出一些通路，问最短的那个通路

思路：

面包店一定和面粉店直接相连,所以贪心从小到大排列通路长度，用vis[ ]记录访问面粉店，那么每条路如果一个被访问一个不被访问，就可以开面包店，更新最小距离就好。

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL __int64
using namespace std;
int vis[100010];
struct node
{
	LL x,y,l;
}a[100010];
bool cmp(node a,node b)//从小到大排列 
{
	return a.l<b.l;
}
int main()
{
	int n,m,k;
	LL x;
	memset(vis,0,sizeof(vis));
	scanf("%d%d%d",&n,&m,&k);
	for(int i=0;i<m;i++)
	{
		scanf("%I64d%I64d%I64d",&a[i].x,&a[i].y,&a[i].l);		
	}
	sort(a,a+m,cmp);
	for(int i=0;i<k;i++)
	{
		scanf("%I64d",&x);
		vis[x]=1;
	}
	LL ans=1e10; 
	for(int i=0;i<m;i++)
	{
		if((vis[a[i].x]&&!vis[a[i].y])||(!vis[a[i].x]&&vis[a[i].y]))
			ans=min(ans,a[i].l);
	}
	if(ans==1e10)	//没有可连通的路 
		printf("-1\n");
	else
		printf("%I64d\n",ans);
	return 0;
}