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
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
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
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int amn=2e5+5,inf=0x3f3f3f3f; 4 int n,k,dis[amn]; 5 int p[amn]; 6 struct node{ 7 int pos,val; 8 bool operator <(const node &a)const {return a.val<val;} 9 }; 10 struct edge{ 11 int to,w,next; 12 }e[amn]; 13 priority_queue<node> que; 14 int head[amn],edgenum=0,vis[amn]; 15 void add(int f,int t,int co) 16 { 17 e[++edgenum].next=head[f]; 18 head[f]=edgenum; 19 e[edgenum].to=t; 20 e[edgenum].w=co; 21 } 22 void Dijkstra(int s) 23 { 24 memset(dis,0x3f,sizeof(dis)); 25 memset(vis,0,sizeof vis); 26 while(que.size())que.pop(); 27 dis[s]=0; 28 node a; 29 a.pos=s,a.val=dis[s]; 30 que.push(a); 31 while(!que.empty()) 32 { 33 node x=que.top();que.pop(); 34 int u = x.pos; 35 if(x.val > dis[u]) continue; 36 vis[u]=true; 37 for(int i=head[u];i!=-1;i=e[i].next) 38 { 39 int v=e[i].to; 40 if(vis[v]) continue; 41 if(dis[v]>e[i].w+dis[u]) 42 { 43 dis[v]=e[i].w + dis[u]; 44 a.pos=v,a.val=dis[v]; 45 que.push(a); 46 } 47 } 48 } 49 } 50 int main(){ 51 memset(head,-1,sizeof head); 52 ios::sync_with_stdio(0); 53 cin>>n>>k; 54 int u,v,ans; 55 for(int i=1;i<n;i++){ 56 cin>>u>>v; 57 add(u,v,-1); 58 add(v,u,-1); 59 } 60 for(int i=1;i<=k;i++){ 61 cin>>p[i]; 62 } 63 Dijkstra(p[1]); 64 int maxn=inf,maxi=0; 65 for(int i=1;i<=k;i++){ 66 if(dis[p[i]]<maxn){ 67 maxn=dis[p[i]]; 68 maxi=i; 69 } 70 } 71 Dijkstra(p[maxi]); 72 ans=inf; 73 for(int i=1;i<=k;i++){ 74 if(dis[p[i]]<ans){ 75 ans=dis[p[i]]; 76 } 77 } 78 ans=-ans; 79 printf("%d ",ans/2+ans%2); 80 }