学步园

二分+BFS：

根据

http://blog.renren.com/blog/240107793/937020122 两个加油站一定在树的直径上。。。

二分长度L ， bfs找到直径的两个端点，在距离直径端点L的地方建加油站，然后bfs检查。。。

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

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

using namespace std;

const int maxn=200200;
const int INF=0x3f3f3f3f;

struct Edge
{
    int to,next;
}edge[maxn*2];

int Adj[maxn],Size;

void init()
{
    memset(Adj,-1,sizeof(Adj)); Size=0;
}

void addedge(int u,int v)
{
    edge[Size].to=v; edge[Size].next=Adj[u]; Adj[u]=Size++;
}

int n;
int dist[maxn],pre[maxn];

int bfs(int u)
{
    queue<int> q;
    q.push(u); dist[u]=0;  pre[u]=0;

    while(!q.empty())
    {
        u=q.front(); q.pop();
        for(int i=Adj[u];~i;i=edge[i].next)
        {
            int v=edge[i].to;
            if(dist[v]>dist[u]+1)
            {
                dist[v]=dist[u]+1;
                pre[v]=u;
                q.push(v);
            }
        }
    }

    return u;
}

bool check(int& s,int& e,int L)
{
    memset(dist,63,sizeof(dist));

    s=bfs(1);
    for(int i=0;i<L;i++)
    {
        s=pre[s];
        if(s==0)
        {
            e=s%n+1;
            return true;
        }
    }

    memset(dist,63,sizeof(dist));
    e=bfs(s);
    for(int i=0;i<L;i++)
    {
        e=pre[e];
        if(e==0)
            return true;
    }

    if(s==e) e=s%n+1;

    bfs(e);

    for(int i=1;i<=n;i++)
        if(dist[i]>L) return false;
    return true;
}

int nextInt()
{
    int ret=0;
    bool ok=false;
    char ch;
    while(ch=getchar())
    {
        if(ch>='0'&&ch<='9')
        {
            ret=ret*10+ch-'0';
            ok=true;
        }
        else if(ok==true) break;
    }
    return ret;
}

int main()
{
    int T_T;
    scanf("%d",&T_T);
    while(T_T--)
    {
        init();
        scanf("%d",&n);
        for(int i=0;i<n-1;i++)
        {
            int u,v;
            u=nextInt(); v=nextInt();
            addedge(u,v); addedge(v,u);
        }
        if(n==2)
        {
            printf("0 1 2\n");
            continue;
        }
        else
        {
            int low=0,high=n,mid,ans;
            int s,e;
            while(low<=high)
            {
                mid=(low+high)/2;
                if(check(s,e,mid)==true)
                {
                    ans=mid;
                    high=mid-1;
                }
                else low=mid+1;
            }
            check(s,e,ans);
            printf("%d %d %d\n",ans,s,e);
        }
    }
    return 0;
}