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
 
Recommend
liuyiding   |   We have carefully selected several similar problems for you:   5065  5064  5062  5061  5060 
 
ここでpath長はいずれも1であり、アクセスするk個の点の数が直径上の点の数より少ない場合、経路長はk−1である.直径を遍歴してもk点に届かない場合は、直径以外の点にアクセスする必要があり、振り返る必要があります

#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
", k - 1);
                continue;
            }
            printf("%d
", maxs + 2 * (k - cnt) );
        }
    }
    return 0;
}