B. Bakery

题目连接:

http://www.codeforces.com/contest/707/problem/B

Description

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 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.

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.

Sample Input

5 4 2

1 2 5

1 2 3

2 3 4

1 4 10

1 5

Sample Output

3

Hint

题意

给你个无向图,有k个特殊的点,问你是否存在一条最短的边,连接特殊的点和不特殊的点。

题解:

把所有边掏出来看一看就好了

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+6;
int n,m,k;
int a[maxn],b[maxn],c[maxn],vis[maxn];
int main()
{
scanf("%d%d%d",&n,&m,&k);
for(int i=1;i<=m;i++)
scanf("%d%d%d",&a[i],&b[i],&c[i]);
for(int i=1;i<=k;i++)
{
int x;
scanf("%d",&x);
vis[x]=1;
}
int ans1=1e9+7;
for(int i=1;i<=m;i++)
{
if(vis[a[i]]+vis[b[i]]==1)
ans1=min(ans1,c[i]);
}
if(ans1==1e9+7)printf("-1\n");
else printf("%d\n",ans1);
}

Codeforces Round #368 (Div. 2) B. Bakery 水题的更多相关文章

  1. Codeforces Round #368 (Div. 2) B. Bakery (模拟)

    Bakery 题目链接: http://codeforces.com/contest/707/problem/B Description Masha wants to open her own bak ...

  2. Codeforces Round #185 (Div. 2) B. Archer 水题

    B. Archer Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/312/problem/B D ...

  3. Codeforces Round #360 (Div. 2) A. Opponents 水题

    A. Opponents 题目连接: http://www.codeforces.com/contest/688/problem/A Description Arya has n opponents ...

  4. Codeforces Round #190 (Div. 2) 水果俩水题

    后天考试,今天做题,我真佩服自己... 这次又只A俩水题... orz各路神犇... 话说这次模拟题挺多... 半个多小时把前面俩水题做完,然后卡C,和往常一样,题目看懂做不出来... A: 算是模拟 ...

  5. Codeforces Round #256 (Div. 2/A)/Codeforces448A_Rewards(水题)解题报告

    对于这道水题本人觉得应该应用贪心算法来解这道题: 下面就贴出本人的代码吧: #include<cstdio> #include<iostream> using namespac ...

  6. Codeforces Round #340 (Div. 2) B. Chocolate 水题

    B. Chocolate 题目连接: http://www.codeforces.com/contest/617/problem/D Descriptionww.co Bob loves everyt ...

  7. Codeforces Round #340 (Div. 2) A. Elephant 水题

    A. Elephant 题目连接: http://www.codeforces.com/contest/617/problem/A Descriptionww.co An elephant decid ...

  8. Codeforces Round #340 (Div. 2) D. Polyline 水题

    D. Polyline 题目连接: http://www.codeforces.com/contest/617/problem/D Descriptionww.co There are three p ...

  9. Codeforces Round #338 (Div. 2) A. Bulbs 水题

    A. Bulbs 题目连接: http://www.codeforces.com/contest/615/problem/A Description Vasya wants to turn on Ch ...

随机推荐

  1. 一些javascript的工具书

    http://pan.baidu.com/s/1jGj9CvO

  2. [转载]CSS Tools: Reset CSS

    http://meyerweb.com/eric/tools/css/reset/ The goal of a reset stylesheet is to reduce browser incons ...

  3. phpStorm 8.0.3 设置

    phpstorm 8 license key Learn Programming===== LICENSE BEGIN =====63758-1204201000000Ryqh0NCC73lpRm!X ...

  4. webp实践的javascript检测方案

    function hasWebp () { // 查看Cookie,如果没有则进行以下逻辑 var img = new Image(); img.onload = handleSupport; img ...

  5. [转]Centos 安装Sublime text 3

    本文简单介绍在CentOS上安装Sublime text 3, 转自:Centos 安装Sublime text 3 Step 1 :建立软件安装目录 # mkdir /opt # cd /opt S ...

  6. Python 入门基础1 --语言介绍

    本节目录: 一.编程语言介绍 二.python解释器介绍 三.安装python解释器 四.运行python程序的两种方式 五.变量 六.后期补充内容 一.编程语言介绍 1.机器语言: 直接用二进制编程 ...

  7. Linux内核启动流程分析(二)【转】

    转自:http://blog.chinaunix.net/uid-25909619-id-3380544.html S3C2410 Linux 2.6.35.7启动分析(第二阶段) 接着上面的分析,第 ...

  8. iptables-25个常用用法【转】

    本文介绍25个常用的iptables用法.如果你对iptables还不甚了解,可以参考上一篇iptables详细教程:基础.架构.清空规则.追加规则.应用实例,看完这篇文章,你就能明白iptables ...

  9. artDialog4.1.7 摘自网络

    jquery.artDialog.source.js学习 1 关键的对象关系 art = jQuery = $ function artDialog() {...} artDialog.fn = ar ...

  10. 浅谈js设计模式之代理模式

    代理模式是一种非常有意义的模式,在生活中可以找到很多代理模式的场景.比如,明星都有经纪人作为代理.如果想请明星来办一场商业演出,只能联系他的经纪人.经纪人会把商业演出的细节和报酬都谈好之后,再把合同交 ...