hdu4607——Park Visit

Park Visit

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2366    Accepted Submission(s): 1051

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≤10 ⁵), 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
   
   
   
   

 

Source

2013 Multi-University Training Contest 1

 

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好在这里的path长度都是1，如果要访问的k个点的数目少于直径上的点的数目，那么路径长度就是k - 1；如果遍历完直径也到不了k个点，那么就需要访问直径以外的点，那么需要走回头路

#include <map>  
#include <set>  
#include <list>  
#include <stack>  
#include <vector>  
#include <queue>  
#include <cmath>  
#include <cstdio>  
#include <cstring>  
#include <iostream>  
#include <algorithm>  
  
using namespace std;  
  
const int N = 100010;  
const int inf = 0x3f3f3f3f;  
  
struct node  
{   
    int next;  
    int to;  
}edge[N << 1];  
  
int n, m;  
int end_p;  
bool vis[N];  
int dist[N];  
int head[N];  
int tot;  
int maxs;  
  
void addedge(int from, int to)  
{  
    edge[tot].to = to;  
    edge[tot].next = head[from];  
    head[from] = tot++;  
}  
  
void build()  
{  
    int u, v;  
    memset( head, -1, sizeof(head) );  
    tot = 0;  
    for (int i = 0; i < n - 1; ++i)  
    {  
        scanf("%d%d", &u, &v); 
        addedge(u, v);  
        addedge(v, u);  
    }  
}  
  
void bfs(int s)  
{  
    queue<int>qu;  
    memset( vis, 0, sizeof(vis) );  
    while ( !qu.empty() )  
    {  
        qu.pop();  
    }  
    qu.push(s);  
    dist[s] = 0;  
    vis[s] = 1;  
    maxs = 0;  
    while ( !qu.empty() )  
    {  
        int u = qu.front();  
        qu.pop();  
        for (int i = head[u]; i != -1; i = edge[i].next)  
        {  
            int v = edge[i].to;  
            if (!vis[v])  
            {  
                vis[v] = 1;  
                dist[v] = dist[u] + 1;  
                qu.push(v);  
                if (maxs < dist[v])  
                {  
                    maxs = dist[v];  
                    end_p = v;  
                } 
            }  
        }  
    }  
}  

int main()
{
    int t, k;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d%d", &n, &m);
        build();
        bfs(1);
        bfs(end_p);
        int cnt = maxs + 1;
        while (m--)
        {
            scanf("%d", &k);
            if(k <= cnt)
            {
                printf("%d\n", k - 1);
                continue;
            }
            printf("%d\n", maxs + 2 * (k - cnt) );
        }
    }
    return 0;
}