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hdu 4607 Park Visit (树的直径)

2024-07-16 18:09:50   219人阅读

Park Visit

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2321    Accepted Submission(s): 1029


Problem Description
Claire and her little friend, ykwd, are travelling in Shevchenko‘s Park! The park is beautiful - but large, indeed. N feature spots in the park are connected by exactly (N-1) undirected paths, and Claire is too tired to visit all of them. After consideration, she decides to visit only K spots among them. She takes out a map of the park, and luckily, finds that there‘re entrances at each feature spot! Claire wants to choose an entrance, and find a way of visit to minimize the distance she has to walk. For convenience, we can assume the length of all paths are 1.
Claire is too tired. Can you help her?
 

Input
An integer T(T≤20) will exist in the first line of input, indicating the number of test cases.
Each test case begins with two integers N and M(1≤N,M≤105), which respectively denotes the number of nodes and queries.
The following (N-1) lines, each with a pair of integers (u,v), describe the tree edges.
The following M lines, each with an integer K(1≤K≤N), describe the queries.
The nodes are labeled from 1 to N.
 

Output
For each query, output the minimum walking distance, one per line.
 

Sample Input
1
4 2
3 2
1 2
4 2
2
4
 

Sample Output
1
4
 

树的直径(Diameter)是指树上的最长简单路。

直径的求法：两遍BFS (or DFS)
任选一点u为起点，对树进行BFS遍历，找出离u最远的点v
以v为起点，再进行BFS遍历，找出离v最远的点w。则v到w的路径长度即为树的直径

求树的直径，任选一个点，从该点开始广搜，最后一个入队的一定是树直径的一端；然后在从这一点开始，再次广搜，则能得到树的直径。然后可以发现，当K小于等于直径Len时，长度为k-1；否则，长度为(k-len)*2+len-1.


#include"stdio.h"
#include"string.h"
#include"queue"
#include"algorithm"
using namespace std;
#define N 100005
struct node
{
    int u,v,next;
}e[N*2];
int a[N];
int head[N],t,len;
bool mark[N];
void add(int u,int v)
{
    e[t].u=u;
    e[t].v=v;
    e[t].next=head[u];
    head[u]=t++;
}
int bfs(int s)
{
    int i,u,v;
    queue<int>q;
    q.push(s);
    memset(mark,false,sizeof(mark));
    mark[s]=true;
    while(!q.empty())
    {
        u=q.front();
        q.pop();
        for(i=head[u];i!=-1;i=e[i].next)
        {
            v=e[i].v;
            if(!mark[v])
            {
                mark[v]=true;
                q.push(v);
            }
        }
    }
    return u;
}
void bfs_len(int s)
{
    int i,v;
    queue<int>q;
    q.push(s);
    memset(mark,false,sizeof(mark));
    mark[s]=true;
    memset(a,0,sizeof(a));
    while(!q.empty())
    {
        s=q.front();
        q.pop();
        len=len>a[s]?len:a[s];
        for(i=head[s];i!=-1;i=e[i].next)
        {
            v=e[i].v;
            if(!mark[v])
            {
                mark[v]=true;
                a[v]=a[s]+1;
                q.push(v);
            }
        }
    }
    return ;
}
int main()
{
    int T,i,u,v,k,n,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        t=0;
        memset(head,-1,sizeof(head));
        for(i=1;i<n;i++)
        {
            scanf("%d%d",&u,&v);
            add(u,v);
            add(v,u);
        }
        len=0;
        u=bfs(1);
        bfs_len(u);
        len++;
        while(m--)
        {
            scanf("%d",&k);
            if(k<=len)
                printf("%d\n",k-1);
            else
                printf("%d\n",(k-len)*2+len-1);
        }
    }
    return 0;
}


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