[原创]codeforces #368 div.2B

2016-08-21 14:51:58 Tabris_ 阅读数：219

博客爬取于2020-06-14 22:43:46
以下为正文

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https://blog.csdn.net/qq_33184171/article/details/52267286

题目链接:http://codeforces.com/contest/707/problem/B
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B. Bakery
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.

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
这里写图片描述

Image illustrates the first sample case. Cities with storage located in and the road representing the answer are darkened.
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题意：
有n个城市 m条双向边 k各仓库分别在a1…ak城市
现在要 找一个没有仓库的城市里找一个城市，使得它到其中一个仓库的距离最短，求这个仓库的距离。
注：-1的情况包括，没有仓库或者全是仓库。

题解：
主要就是不能用矩阵来存储图就行了 鄙人用的vector

代码
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# include <bits/stdc++.h>
using namespace std;

# define LL long long int
# define INF 0x1f1f1f1f
# define pb push_back

/*************************************/

const int MOD = 1e9+7;
const int M = 1e5+10;
struct node
{
    int a,l;
};
vector<node>E[M];
int has[M];


int main()
{
    ios::sync_with_stdio(false);
    int n,m,k;
    memset(has,0,sizeof(has));
    for(int i=1; i<=M; i++)
        E[i].clear();

    cin>>n>>m>>k;
    int a,b,c;

    for(int i=0; i<m; i++)
    {
        cin>>a>>b>>c;
        E[a].pb(node {b,c});
        E[b].pb(node {a,c});
    }

    int x;
    int num=0,flag=0;
    int mini=2000000000;

    for(int i=0; i<k; i++)
    {
        cin>>x;
        if(!has[x])     has[x]=1,num++;
    }

    if(num==0||num==n) flag=1;
     //printf("%d\n",flag);
    for(int i=1; i<=n&&!flag; i++)
        if(!has[i])
        {
            for(int j=0; j<E[i].size(); j++)
                if(has[E[i][j].a])
                    mini=min(mini,E[i][j].l);
        }


    if(flag==1||mini==2000000000) puts("-1");
    else              printf("%d\n",mini);
    return 0;
}