【多校训练】hdu 6166 Senior Pan 最短路径 Dijkstra

Problem Description

Senior Pan fails in his discrete math exam again. So he asks Master ZKC to give him graph theory problems everyday.
The task is simple : ZKC will give Pan a directed graph every time, and selects some nodes from that graph, you can calculate the minimum distance of every pair of nodes chosen in these nodes and now ZKC only cares about the minimum among them. That is still too hard for poor Pan, so he asks you for help.

 

Input

The first line contains one integer T, represents the number of Test Cases.1≤T≤5.Then T Test Cases, for each Test Cases, the first line contains two integers n,m representing the number of nodes and the number of edges.1≤n,m≤100000
Then m lines follow. Each line contains three integers xi,yi representing an edge, and vi representing its length.1≤xi,yi≤n,1≤vi≤100000
Then one line contains one integer K, the number of nodes that Master Dong selects out.1≤K≤n
The following line contains K unique integers ai, the nodes that Master Dong selects out.1≤ai≤n,ai!=aj

 

Output

For every Test Case, output one integer: the answer

 

Sample Input

15 61 2 12 3 33 1 32 5 12 4 24 3 131 3 5

 

Sample Output

Case #1: 2

题意：

一个有向图，选其中k个点，求两点间距离的最小值。

思路：

我们只要求出任意点对的最小值即可，但是暴力做肯定会超时。把k个点分成两个集合，用dijkstra求两个集合间的最小值==求两个集合间的所有点对的最小值。

由于任意两点的二进制不同，我们可以枚举二进制的位，将k分成两个集合。只需要logn次dijkstra就能得到所有点对间的最小值。

////  main.cpp//  1006////  Created by zc on 2017/8/23.//  Copyright © 2017年 zc. All rights reserved.//#include <iostream>#include<cstdio>#include<queue>#include<cstring>#include<cmath>#include<algorithm>#define ll long longusing namespace std;const int M=440000;const int N=110000;const ll INF=0x3f3f3f3f3f3f3f3f;int n,m,cnt,head[N],k,a[N],v[N];ll d[N];struct node{    int next,v;    ll c;}e[M];struct node2{    int u;    ll c;    friend bool operator< (node2 a ,node2 b)    {        return a.c>b.c;    }};priority_queue<node2> q;void add(int x,int y,ll c){    e[cnt].c=c;    e[cnt].v=y;    e[cnt].next=head[x];    head[x]=cnt++;}void init(){    memset(v,0,sizeof(v));    memset(d,0x3f,sizeof(d));    while(!q.empty())   q.pop();}ll dijkstra(){    while(!q.empty())    {        node2 t=q.top();        q.pop();        int u=t.u;        ll c=t.c;        if(v[u])    return c;        for(int i=head[u];i+1;i=e[i].next)        {            int y=e[i].v;            if(d[y]>d[u]+e[i].c)            {                d[y]=c+e[i].c;                q.push(node2{y,d[y]});            }        }    }    return INF;}ll solve(){    ll ans=INF;    for(int i=0;i<20;i++)    {        init();        for(int j=0;j<k;j++)        {            if(a[j]&(1<<i))            {                q.push(node2{a[j],0});                d[a[j]]=0;            }            else            {                v[a[j]]=1;            }        }        ans=min(ans,dijkstra());    }    return ans;}int main(int argc, const char * argv[]) {    int T,kase=0;    scanf("%d",&T);    while(T--)    {        cnt=0;        memset(head,-1,sizeof(head));        scanf("%d%d",&n,&m);        for(int i=0;i<m;i++)        {            int x,y;            ll c;            scanf("%d%d%lld",&x,&y,&c);            add(x,y,c);        }        scanf("%d",&k);        for(int i=0;i<k;i++)    scanf("%d",&a[i]);        printf("Case #%d: %lld\n",++kase,solve());    }}