2019牛客多校第四场 A meeting

链接：https://ac.nowcoder.com/acm/contest/884/A
来源：牛客网

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 524288K，其他语言1048576K
64bit IO Format: %lld

题目描述

A new city has just been built. There're nnn interesting places numbered by positive numbers from 111 to nnn.

In order to save resources, only exactly n−1n-1n−1 roads are built to connect these nnn interesting places. Each road connects two places and it takes 1 second to travel between the endpoints of any road.

There is one person in each of the places numbered x1,x2…xkx_1,x_2 \ldots x_kx1,x2…xk and they've decided to meet at one place to have a meal. They wonder what's the minimal time needed for them to meet in such a place. (The time required is the maximum time for each person to get to that place.)

输入描述:

First line two positive integers, n,kn,kn,k - the number of places and persons.

For each the following n−1n-1n−1 lines, there're two integers a,ba,ba,b that stand for a road connecting place aaa and bbb. It's guaranteed that these roads connected all nnn places.

On the following line there're kkk different positive integers x1,x2…xkx_1,x_2 \ldots x_kx1,x2…xk separated by spaces. These are the numbers of places the persons are at.

输出描述:

A non-negative integer - the minimal time for persons to meet together.

示例1

输入

复制

输出

复制

说明

They can meet at place 1 or 3.
1<=n<=1e5

题意

给出一颗树和一些点,一个点到另一个点的距离为1,求这些点在树上最大距离的一半

思路
要求图上的最大距离可先任意取一个点求最大距离,再从最大距离的这个点再找一次最大距离(因为重新求的最大距离必定大于或等于原先的最大距离)
这里求最大距离用了边权取负再dijkstra,把求出的最小负距离取反就得到最大正距离

代码

 #include<bits/stdc++.h>

 using namespace std;

 const int amn=2e5+,inf=0x3f3f3f3f;

 int n,k,dis[amn];

 int p[amn];

 struct node{

     int pos,val;

     bool operator <(const node &a)const {return a.val<val;}

 };

 struct edge{

     int to,w,next;

 }e[amn];

 priority_queue<node> que;

 int head[amn],edgenum=,vis[amn];

 void add(int f,int t,int co)

     {

         e[++edgenum].next=head[f];

         head[f]=edgenum;

         e[edgenum].to=t;

         e[edgenum].w=co;

     }

 void Dijkstra(int s)

     {

         memset(dis,0x3f,sizeof(dis));

         memset(vis,,sizeof vis);

         while(que.size())que.pop();

         dis[s]=;

         node a;

         a.pos=s,a.val=dis[s];

         que.push(a);

         while(!que.empty())

             {

                 node x=que.top();que.pop();

                 int u = x.pos;

                 if(x.val > dis[u]) continue;

                 vis[u]=true;

                 for(int i=head[u];i!=-;i=e[i].next)

                     {

                         int v=e[i].to;

                         if(vis[v])    continue;

                         if(dis[v]>e[i].w+dis[u])

                             {

                                 dis[v]=e[i].w + dis[u];

                                 a.pos=v,a.val=dis[v];

                                 que.push(a);

                             }

                     }

             }

     }

 int main(){

     memset(head,-,sizeof head);

     ios::sync_with_stdio();

     cin>>n>>k;

     int u,v,ans;

     for(int i=;i<n;i++){

         cin>>u>>v;

         add(u,v,-);

         add(v,u,-);

     }

     for(int i=;i<=k;i++){

         cin>>p[i];

     }

     Dijkstra(p[]);

     int maxn=inf,maxi=;

     for(int i=;i<=k;i++){

         if(dis[p[i]]<maxn){

             maxn=dis[p[i]];

             maxi=i;

         }

     }

     Dijkstra(p[maxi]);

     ans=inf;

     for(int i=;i<=k;i++){

         if(dis[p[i]]<ans){

             ans=dis[p[i]];

         }

     }

     ans=-ans;

     printf("%d\n",ans/+ans%);

 }