zoj 3820 Building Fire Stations （二分+树的直径）

Building Fire Stations

Time Limit: 5 Seconds
Memory Limit: 131072 KB Special Judge

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

Sample Output

1 1 2

1 2 4

题意：

给你一颗树。选两个点做消防站，使得全部点到近期消防站的最大距离最小。

思路：

能够yy到两个点选在树的直径上，于是能够二分答案，答案为 mid 的时候两个端点就应该选 st+mid 和 ed-mid 这两个点（st、ed是树的直径的端点），然后枚举中间的那些点是否满足条件来检验就够了。

代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

#include <vector>

#include <queue>

#include <algorithm>

#define maxn 200005

#define INF 0x3f3f3f3f

typedef long long ll;

using namespace std;

int n,m,ans;

int st,ed,dd,resu,resv;

vector<int>edge[maxn];

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

int indst[maxn],inded[maxn];

bool vis[maxn],app[maxn];

int bfs(int sx)

{

    int i,j,u,v,res;

    queue<int>q;

    memset(dist,-1,sizeof(dist));

    memset(pre,0,sizeof(pre));

    dist[sx]=dd=0;

    q.push(sx);

    while(!q.empty())

    {

        u=q.front();

        if(dist[u]>=dd)

        {

            res=u; dd=dist[u];

        }

        q.pop();

        for(i=0;i<edge[u].size();i++)

        {

            v=edge[u][i];

            if(dist[v]==-1)

            {

                pre[v]=u;

                dist[v]=dist[u]+1;

                q.push(v);

            }

        }

    }

    return res;

}

int bfs1(int sx)

{

    int i,j,u,v,res=0;

    queue<int>q;

    city[sx]=0;

    q.push(sx);

    while(!q.empty())

    {

        u=q.front();

        res=max(res,city[u]);

        q.pop();

        for(i=0;i<edge[u].size();i++)

        {

            v=edge[u][i];

            if(!vis[v]&&!app[v])

            {

                app[v]=1;

                city[v]=city[u]+1;

                q.push(v);

            }

        }

    }

    return res;

}

bool isok(int mid)

{

    int i,j,u,v,tu=indst[mid],tv=inded[mid],gap=dist[ed]-2*mid;

    //printf("mid:%d tu:%d tv:%d gap:%d\n",mid,tu,tv,gap);

    u=tv;

    int num=0;

    while(u!=tu)

    {

        if(city[u]+min(num,gap-num)>mid) return false ;

        u=pre[u];

        num++;

    }

    return true ;

}

void solve()

{

    int i,j,u,v,num=0;

    st=bfs(1);

    ed=bfs(st);

   // printf("st:%d ed:%d\n",st,ed);

    memset(vis,0,sizeof(vis));

    u=ed;

    while(u)

    {

        inded[num]=u; indst[dist[ed]-num]=u;

        num++;

        vis[u]=1;

        u=pre[u];

    }

    memset(app,0,sizeof(app));

    u=ed;

    while(u)

    {

        city[u]=bfs1(u);

        //printf("u:%d city:%d\n",u,city[u]);

        u=pre[u];

    }

    int le=0,ri=dist[ed]/2,mid;

    while(le<=ri)

    {

        mid=(le+ri)>>1;

        if(isok(mid))

        {

            ans=mid; resu=indst[mid]; resv=inded[mid];

            if(resu==resv) resu=pre[resu];

            ri=mid-1;

        }

        else le=mid+1;

    }

    printf("%d %d %d\n",ans,resu,resv);

}

int main()

{

    int i,j,t;

    scanf("%d",&t);

    while(t--)

    {

        scanf("%d",&n);

        for(i=1;i<=n;i++) edge[i].clear();

        int u,v;

        for(i=1;i<n;i++)

        {

            scanf("%d%d",&u,&v);

            edge[u].push_back(v);

            edge[v].push_back(u);

        }

        solve();

    }

    return 0;

}

1 1 2

1 2 4