Title description
A new city has just been built. There're
interesting places numbered by positive numbers from 1 to
In order to save resources, only exactly
roads are built to connect these
There is one person in each of the places numbered
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.)
Enter description
First line two positive integers, n,k - the number of places and persons.
For each the following n−1 lines, there're two integers a,b that stand for a road connecting place a and b. It's guaranteed that these roads connected all n places.
On the following line there're k different positive integers
Output description
A non-negative integer - the minimal time for persons to meet together.
Example input
4 2
1 2
3 1
3 4
2 4
Sample output
2
Explanation
They can meet at place 1 or 3.
Remarks
The main idea:
In a tree, there are multiple nodes, and someone asks to select a node so that everyone can meet at this node. The meeting time is defined as the longest time it takes for everyone to reach the node. The minimum meeting time is required.
analysis:
Consider the two people who are farthest apart, then the midpoint of the two paths is the meeting point of all people, which can ensure the minimum meeting time. This is similar to finding the diameter of a tree. The result is that the diameter of the tree is divided by 2 and rounded up. The BFS is used twice to find the diameter, but it should be noted that only the depth of the node of the person is used to update the diameter during the traversal, so the maximum distance between the nodes of the person is obtained.
See the code for specific explanation.
#include<stdio.h>
#include<string.h>
#include<queue>
#define MAX 200005
using namespace std;
int head[MAX];
int vis[MAX];//在BFS中，标记当前节点是否已经用过
int dis[MAX];//记录最长距离（即节点的深度）
int signal[MAX];//标志哪些点有人
int n,m,ans;
int sum;//记录有人的节点之间最长路径的长度
int lst;//记录最长路径的终点节点
struct node
{
int u,v,w;
int next;
}edge[MAX];
void add(int u,int v,int w)//向邻接表中加边
{
edge[ans].u=u;
edge[ans].v=v;
edge[ans].w=w;
edge[ans].next=head[u];
head[u]=ans++;
}
void getmap()
{
int i,j;
int a,b,c;
ans=0;
memset(head,-1,sizeof(head));
while(m--)
{
scanf("%d%d",&a,&b);
add(a,b,1);
add(b,a,1);
}
}
void bfs(int beg)
{
queue<int>q;
memset(dis,0,sizeof(dis));
memset(vis,0,sizeof(vis));
int i,j;
while(!q.empty())
q.pop();
lst=beg;//终点初始化为起点
sum=0;
vis[beg]=1;
q.push(beg);
int top;
while(!q.empty())
{
top=q.front();
q.pop();
for(i=head[top];i!=-1;i=edge[i].next)
{
if(!vis[edge[i].v])
{
dis[edge[i].v]=dis[top]+edge[i].w;//更新节点深度
vis[edge[i].v]=1;
q.push(edge[i].v);
if(signal[edge[i].v]&&sum<dis[edge[i].v])//只有当该节点有人，才会用该节点的深度去更新最长路径的值
{
sum=dis[edge[i].v];
lst=edge[i].v;
}
}
}
}
}
int main()
{
int n,k,i;
scanf("%d%d",&n,&k);
m=n-1;
getmap();
int a;
for(i=1;i<=k;i++)
{
scanf("%d",&a);
signal[a]=1;
}
bfs(a);//从某一个有人的节点开始，搜索最长路径的一个端点
bfs(lst);//搜索另一个端点
printf("%d\n",(sum+1)/2);//最长路径的一半上取整
return 0;
}
