MUTC2013 H-Park Visit-hdu4607

Park Visit

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 85    Accepted Submission(s): 34

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

 

Source

2013 Multi-University Training Contest 1

 

Recommend

liuyiding

 

求出树的直径r，当k<=r时，ans=k-1；当k>r时，ans=(k-r)*2+r-1;

-----

第一次尝试使用超DIO的模板。

/** head-file **/

#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <algorithm>

/** define-for **/

#define REP(i, n) for (int i=0;i<int(n);++i)
#define FOR(i, a, b) for (int i=int(a);i<int(b);++i)
#define DWN(i, b, a) for (int i=int(b-1);i>=int(a);--i)
#define REP_1(i, n) for (int i=1;i<=int(n);++i)
#define FOR_1(i, a, b) for (int i=int(a);i<=int(b);++i)
#define DWN_1(i, b, a) for (int i=int(b);i>=int(a);--i)
#define REP_N(i, n) for (i=0;i<int(n);++i)
#define FOR_N(i, a, b) for (i=int(a);i<int(b);++i)
#define DWN_N(i, b, a) for (i=int(b-1);i>=int(a);--i)
#define REP_1_N(i, n) for (i=1;i<=int(n);++i)
#define FOR_1_N(i, a, b) for (i=int(a);i<=int(b);++i)
#define DWN_1_N(i, b, a) for (i=int(b);i>=int(a);--i)

/** define-useful **/

#define clr(x,a) memset(x,a,sizeof(x))
#define sz(x) int(x.size())
#define see(x) cerr<<#x<<" "<<x<<endl
#define se(x) cerr<<" "<<x
#define pb push_back
#define mp make_pair

/** test **/

#define Display(A, n, m) {                      
    REP(i, n){                                  
        REP(j, m) cout << A[i][j] << " ";       
        cout << endl;                           
    }                                           
}

#define Display_1(A, n, m) {                    
    REP_1(i, n){                                
        REP_1(j, m) cout << A[i][j] << " ";     
        cout << endl;                           
    }                                           
}

using namespace std;

/** typedef **/

typedef long long LL;

/** Add - On **/

const int direct4[4][2]= { {0,1},{1,0},{0,-1},{-1,0} };
const int direct8[8][2]= { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int direct3[6][3]= { {1,0,0},{0,1,0},{0,0,1},{-1,0,0},{0,-1,0},{0,0,-1} };

const int MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long INFF = 1LL << 60;
const double EPS = 1e-9;
const double OO = 1e15;
const double PI = acos(-1.0); //M_PI;

const int maxn=111111;
const int maxm=511111;
int n,m;

struct EDGENODE
{
    int to;
    int w;
    int next;
};

int head[maxn];
EDGENODE edges[maxm];
int edge;
void init()
{
    clr(head,-1);
    edge=0;
}
void addedge(int u,int v,int c=0)
{
    edges[edge].w=c,edges[edge].to=v,edges[edge].next=head[u],head[u]=edge++;
    edges[edge].w=c,edges[edge].to=u,edges[edge].next=head[v],head[v]=edge++;
}

bool vis[maxn];
int len[maxn];

void dfs(int u,int pa)
{
    len[u]=1;
    if (pa!=-1) len[u]+=len[pa];
    for (int i=head[u]; i!=-1; i=edges[i].next)
    {
        int v=edges[i].to;
        if (v!=pa&&!vis[v])
        {
            dfs(v,u);
        }
    }
}

int get_tree_len()
{
    int ret=0,t;
    clr(vis,0);
    dfs(1,-1);
    REP_1(i,n) if (len[i]>ret)
    {
        ret=len[i];
        t=i;
    }
    clr(vis,0);
    dfs(t,-1);
    REP_1(i,n) ret=max(ret,len[i]);
    return ret;
}

int main()
{
    int T;
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d%d",&n,&m);
        init();
        REP(i,n-1)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            addedge(u,v);
        }
        int r=get_tree_len();
        //see(r);
        REP(i,m)
        {
            int k;
            scanf("%d",&k);
            if (k<=r) printf("%d
",k-1);
            else  printf("%d
",(k-r)*2+r-1);
        }
    }
    return 0;
}