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. bzoj千题计划228:bzoj2095: [Poi2010]Bridges

    http://www.lydsy.com/JudgeOnline/problem.php?id=2095 二分答案,判断是否存在混合图的欧拉回路 如果只有一个方向的风力<=mid,这条边就是单向 ...

  2. 考研:操作系统:进程同步—信号量实现同步互斥(PV操作)

    进程互斥的硬件实现方法

  3. css 基础2

    1.内部样式表: 2.行内样式表:在标签内写style,适合style 比较少的情况 3.外部样式表(外联式): 4.html标签可以分为:块级标签,h1~h6,div ,p,ul,ol,li,div ...

  4. 用python处理文本,本地文件系统以及使用数据库的知识基础

    主要是想通过python之流的脚本语言来进行文件系统的遍历,处理文本以及使用简易数据库的操作. 本文基于陈皓的:<程序员技术练级攻略> 一.Python csv 对于电子表格和数据库导出文 ...

  5. CentOS 无法通过 yum 安装新版 nodejs 解决办法(安装的还是老版的)

    官网安装说明:CentOS 安装 nodejs 第一步: curl --silent --location https://rpm.nodesource.com/setup_10.x | sudo b ...

  6. 洛谷 P4389: 付公主的背包

    题目传送门:洛谷 P4389. 题意简述: 有 \(n\) 个物品,每个物品都有无限多,第 \(i\) 个物品的体积为 \(v_i\)(\(v_i\le m\)). 问用这些物品恰好装满容量为 \(i ...

  7. Linux下USB转串口的驱动【转】

    转自:http://www.linuxidc.com/Linux/2011-02/32218.htm Linux发行版自带usb to serial驱动,以模块方式编译驱动,在内核源代码目录下运行Ma ...

  8. 使用pt-table-checksum校验MySQL主从复制【转】

    pt-table-checksum是一个基于MySQL数据库主从架构在线数据一致性校验工具.其工作原理在主库上运行, 通过对同步的表在主从段执行checksum, 从而判断数据是否一致.在校验完毕时, ...

  9. SecureCRT中常用linux命令 -《转载》

    常用命令: 一.ls 只列出文件名 (相当于dir,dir也可以使用) -A:列出所有文件,包含隐藏文 件. -l:列表形式,包含文件的绝大部分属性. -R:递归显示. --help:此命令的帮助. ...

  10. ConvertUtils.register的作用

    BeanUtils的populate方法或者getProperty,setProperty方法其实都会调用convert进行转换 但Converter只支持一些基本的类型,甚至连Java.util.D ...