ZOJ 3826 Hierarchical Notation(2014 牡丹江 B,树的直径+二分)

最新推荐文章于 2018-01-26 18:29:51 发布

原创最新推荐文章于 2018-01-26 18:29:51 发布 · 614 阅读

0 ·

CC 4.0 BY-SA版权

在Marjar大学的美丽和平静环境中，通过构建一条连接所有建筑物的唯一路径，并利用二分搜索算法，找出最佳的两座建筑作为消防站的位置，以确保在发生火灾时，消防员能够尽快到达现场。

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 andN - 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 andY_i. That means there is a road connecting building X_i and buildingY_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

题目大意：
给你一棵树，每条边边权都是1，找2个不同的哨兵点，使得所有的点到最近的哨兵点的距离的最大值最小。
解题思路：
由于是求最大值最小问题，那么一般会想到二分答案求解。
那么问题来了，二分一个答案之后如果安放这两个点的位置。考虑到树的直径。由于是最大值，所以这两个点到两个直径端的距离 应该等于二分的答案X.然后在判断其他点
到这2个点的距离的最大值。如果小于等于X的话 说明X满足答案。否则X不满足 那么再小的也不满足。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <queue>
#define MOD 1000000007
#define eps 1e-8
#define maxn 200000+10
#define maxm 400000+10


using namespace std;

struct Edge
{
    int from,to,next;
}es[maxm];
int cnt,p[maxn];
int pre[maxn];

void add(int x,int y)
{
    es[cnt].from = x;
    es[cnt].to = y;
    es[cnt].next=  p[x];
    p[x] = cnt++;
}

int dis[maxn];
int used[maxn];

void init()
{
    memset(p,-1,sizeof p);
   // memset(used,0,sizeof used);
    cnt = 0;
}
int n,st,en;
int maxlen;

void bfs(int u)
{
    memset(used,0,sizeof(used));
    memset(dis,0x7F,sizeof(dis));
    dis[u]=0;
    used[u]=1;
    queue<int>q;
    q.push(u);
    while(!q.empty())
    {
        u=q.front();
        q.pop();
        if(dis[u]>maxlen)
        {
            maxlen=dis[u];
            st=u;
        }
        for(int i = p[u];~i;i = es[i].next)
        {
            int v = es[i].to;
            if(used[v]) continue;
            used[v] = 1;
            dis[v] =dis[u]+1;
            q.push(v);
        }
    }
}

void bfs1(int u)
{
    memset(used,0,sizeof(used));
    memset(dis,0x7F,sizeof(dis));
    dis[u]=0;
    used[u]=1;
    queue<int>q;
    q.push(u);
    while(!q.empty())
    {
        u=q.front();
        q.pop();
        if(dis[u]>maxlen)
        {
            maxlen=dis[u];
            en=u;
        }
        for(int i = p[u];~i;i = es[i].next)
        {
            int v = es[i].to;
            if(used[v]) continue;
            used[v] = 1;
            dis[v] =dis[u]+1;
            pre[v]=u;
            q.push(v);
        }
    }
}
void bfs2(int u)
{
    memset(used,0,sizeof(used));
    dis[u]=0;
    used[u]=1;
    queue<int>q;
    q.push(u);
    while(!q.empty())
    {
        u=q.front();
        q.pop();
        for(int i = p[u];~i;i = es[i].next)
        {
            int v = es[i].to;
            if(used[v]) continue;
            used[v] = 1;
            dis[v] = min(dis[v],dis[u]+1);
            q.push(v);
        }
    }
}

int l1,l2;

bool judge(int x)
{
    l1 = en;
    l2 = en;
    int t = x;
    if(t<0) return 0;
    while(t--)
    {
        l1 = pre[l1];
    }
    t = maxlen -x ;
    if(t<0) return 0;
    while(t--)
    {
        l2 = pre[l2];
    }
    if(l2 == l1)
        l2 = pre[l2];
    memset(dis,0x7F,sizeof dis);
    dis[l1]=0;

    bfs2(l1);
    dis[l2]=0;
    bfs2(l2);
   // if(x==1)
    //    for(int i=1;i<=n;i++) cout<<dis[i]<<' ';
   // cout<<endl;
    for(int i=1;i<=n;i++)
        if(dis[i]>x) return 0;
    //if(!dfs2(l1,-1,x)) return 0;
   // if(!dfs2(l2,-1,x)) return 0;
    return 1;
}

int main()
{

    int T;
    scanf("%d",&T);
    for(int ks = 1;ks <= T;ks++)
    {
        scanf("%d",&n);
        init();
        for(int i = 0;i < n-1;i++)
        {
            int x,y;
            scanf("%d %d",&x,&y);
            add(x,y);
            add(y,x);
        }
        maxlen = 0;
        memset(dis,0,sizeof dis);
        memset(used,0,sizeof used);
        bfs(1);
        maxlen = 0;
        memset(dis,0,sizeof dis);
        memset(used,0,sizeof used);
        bfs1(st);

        int ans = 0;
        int ans1,ans2;
        int l = 0,r= maxlen/2;
        while(l <= r)
        {
            int mid = (l+r)>>1;
            if(judge(mid))
            {
                ans = mid;
                r= mid-1;
                ans1 = l1;
                ans2 = l2;
            }
            else l = mid+1;
          //  cout<<"fsd"<<endl;
        }
        printf("%d %d %d\n",ans,ans1,ans2);
    }
}