zoj 3820 Building Fire Stations 树的中心

Building Fire Stations

Time Limit: 1 Sec  

Memory Limit: 256 MB

题目连接

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3820

Description

Marjar University is a beautiful and peaceful place. There are N buildings and N - 1 bidirectional roads in the campus. These buildings are connected by roads in such a way that there is exactly one path between any two buildings. By coincidence, the length of each road is 1 unit.

To ensure the campus security, Edward, the headmaster of Marjar University, plans to setup two fire stations in two different buildings so that firefighters are able to arrive at the scene of the fire as soon as possible whenever fires occur. That means the longest distance between a building and its nearest fire station should be as short as possible.

As a clever and diligent student in Marjar University, you are asked to write a program to complete the plan. Please find out two proper buildings to setup the fire stations.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains an integer N (2 <= N <= 200000).

For the next N - 1 lines, each line contains two integers X_i and Y_i. That means there is a road connecting building X_i and building Y_i(indexes are 1-based).

Output

For each test case, output three integers. The first one is the minimal longest distance between a building and its nearest fire station. The next two integers are the indexes of the two buildings selected to build the fire stations.

If there are multiple solutions, any one will be acceptable.

Sample Input

2
4
1 2
1 3
1 4
5
1 2
2 3
3 4
4 5

 

Sample Output

1 1 2
1 2 4

HINT

 

题意

给你一棵树，然后找俩点，使得其他点到这俩点的最短距离最长边最小

题解：

我们是找的直径，然后按着直径剪开，然后变成了两棵树，然后取拆出来的两棵树的中心就好了

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <queue>

using namespace std;

const int N=200010;
int ans1,ans2,l,n,T;
int pre[N*2],to[N*2],nxt[N*2],cnt,d[N],s1,s2,s3,s4,s5,s6,ss,tt,ans,fa[N];

void makeedge(int x,int y)
{
    to[cnt]=y;nxt[cnt]=pre[x];pre[x]=cnt++;
    to[cnt]=x;nxt[cnt]=pre[y];pre[y]=cnt++;
}

int bfs1(int s,int no)
{
    int z=s;
    queue<int> q;
    while(!q.empty()) q.pop();
    for(int i=0;i<=n;i++) d[i]=0;
    q.push(s);
    d[s]=1;
    while(!q.empty())
    {
        int x=q.front();
        q.pop();
        for(int p=pre[x];p!=-1;p=nxt[p])
        {
            int y=to[p];
            if(d[y]||y==no) continue;
            d[y]=d[x]+1;
            z=y;
            fa[y]=x;
            q.push(y);
        }
    }
    return z;
}
int dfs(int x,int no,int dep)
{
    if(d[x]==dep)
    {
        if(no==0)tt=fa[x];
        return x;
    }
    return dfs(fa[x],no,dep);
}

int main()
{
    scanf("%d",&T);
    while(T--)
    {
        memset(pre,-1,sizeof(pre));
        cnt=0;
        scanf("%d",&n);
        for(int i=1;i<n;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            makeedge(x,y);
        }
        s1=bfs1(1,0);
        s2=bfs1(s1,0);
        ss=dfs(s2,0,(d[s2]+2)/2);
        s3=bfs1(ss,tt);
        s4=bfs1(s3,tt);
        ans1=dfs(s4,tt,(d[s4]+1)/2);
        ans=max(d[ans1]-1,d[s4]-d[ans1]);
        s5=bfs1(tt,ss);
        s6=bfs1(s5,ss);
        ans2=dfs(s6,ss,(d[s6]+1)/2);
        ans=max(ans,max(d[ans2]-1,d[s6]-d[ans2]));
        printf("%d %d %d\n",ans,ans1,ans2);
    }
}