题目描述
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.)
输入描述
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
输出描述
A non-negative integer - the minimal time for persons to meet together.
示例输入
4 2
1 2
3 1
3 4
2 4
示例输出
2
说明
They can meet at place 1 or 3.
备注
题目大意：
在一棵树上，有多个节点有人，要求选择一个节点，让所有人在这个节点相遇。相遇的时间规定为所有人中到达该节点所用的最长时间。要求给出最小的相遇时间。
分析：
考虑相距最远的两个人，那么两者路径的中点就是所有人的相遇点，这样能保证最小的相遇时间。这就类似于求树的直径，结果就是树的直径除以2然后向上取整。这里用两遍BFS求直径，但要注意在遍历的过程中只用有人的节点的深度去更新直径，这样求出来的就是有人的节点之间的最远距离。
具体解释见代码。
#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;
}
