hdu 5927 Auxiliary Set dfs

Auxiliary Set

Time Limit: 9000/4500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 318    Accepted Submission(s): 93

Problem Description

Given a rooted tree with n vertices, some of the vertices are important.

An auxiliary set is a set containing vertices satisfying at least one of the two conditions：

∙ It is an important vertex
∙ It is the least common ancestor of two different important vertices.

You are given a tree with n vertices (1 is the root) and q queries.

Each query is a set of nodes which indicates the  unimportant vertices in the tree. Answer the size (i.e. number of vertices) of the auxiliary set for each query.

 

Input

The first line contains only one integer T ( T≤1000 ), which indicates the number of test cases.

For each test case, the first line contains two integers n ( 1≤n≤100000 ), q ( 0≤q≤100000 ).

In the following n -1 lines, the i-th line contains two integers  ui,vi(1≤ui,vi≤n)  indicating there is an edge between  ui i and  vi  in the tree.

In the next q lines, the i-th line first comes with an integer  mi(1≤mi≤100000)  indicating the number of vertices in the query set.Then comes with mi different integers, indicating the nodes in the query set.

It is guaranteed that  ∑qi=1mi≤100000 .

It is also guaranteed that the number of test cases in which  n≥1000   or  ∑qi=1mi≥1000  is no more than 10.

 

Output

For each test case, first output one line "Case #x:", where x is the case number (starting from 1).

Then q lines follow, i-th line contains an integer indicating the size of the auxiliary set for each query. 

 

Sample Input

  
  

   
   1
6 3
6 4
2 5
5 4
1 5
5 3
3 1 2 3
1 5
3 3 1 4
  
  

 

Sample Output

  
  

   
   Case #1:
3
6
3

   
   

    
    

     
     Hint
    
    


    
    

For the query {1，2, 3}:
•node 4, 5, 6 are important nodes For the query {5}：
•node 1，2, 3, 4, 6 are important nodes
•node 5 is the lea of node 4 and node 3 For the query {3, 1，4}:
• node 2, 5, 6 are important nodes 

   
   
   
    
  
  
  
  

   
   

  
  
  
  

   
   

  
  
  
  

   
   树上的dfs，找到每个重要结点或者两个重要节点的lca，求个数，但是初始化真是服气。。。
   
   It is also guaranteed that the number of test cases in which 
   
   
   
   
    
    
     
     n≥1000
    
    
   
     or 
   
   
   
   
    
    
     
     ∑qi=1mi≥1000
    
    
   
    is no more than 10.这句话说明不能每次更新询问时都重新初始化，大部分时候m<1000，所以我们更新是每次更新m个非重要点之后查询结束后再更新回来，会将复杂度从o(n^2)降到o(n*m);
  
  
  
  

   
   吃一鉴长一智
  
  
  
  

   
   

  
  
  
  

   
   
   
   
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=100010;
int n,m,q;
int im[maxn];
int cnt;
int qn[maxn];
int fa[maxn];
struct Node
{
    int to;
    int next;
};
struct Node edges[2*maxn];
int head[maxn];
int k;
void AddNode(int from, int to)
{
    edges[k].to=to;
    edges[k].next=head[from];
    head[from]=k++;
}
int dfs(int u,int p)
{
    if(im[u]==1)
        return 1;
    else if(im[u]==-1)
        return 0;
    int res=0;
    for(int i=head[u]; i!=-1; i=edges[i].next)
    {
        int v=edges[i].to;
        if(v==p)
            continue;
        res+=dfs(v,u);
        if(res==2)
            break;
    }
    if(res==1)
    {
        im[u]=1;
        return 1;
    }
    if(res==2)
    {
        im[u]=1;
        cnt++;
        return 1;
    }
    im[u]=-1;
    return 0;
}
void solve(int u)
{
    int res=0;
    for(int i=head[u]; i!=-1; i=edges[i].next)
    {
        int v=edges[i].to;
        if(v==fa[u])
            continue;
        res+=dfs(v,u);
        if(res==2)
        {
            im[u]=1;
            cnt++;
            return;
        }
    }
    if(res==0)
        im[u]=-1;
    else if(res)
        im[u]=1;
}
void find_fa(int u,int p)
{
    fa[u]=p;
    for(int i=head[u]; i!=-1; i=edges[i].next)
    {
        int v=edges[i].to;
        if(v==p)
            continue;
        find_fa(v,u);
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    int cns=0;
    while(t--)
    {
        scanf("%d%d",&n,&q);
        k=0;
        memset(head,-1,sizeof(head));
        for(int i=1; i<=n; i++)
            im[i]=1;
        for(int i=0; i<n-1; i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            AddNode(x,y);
            AddNode(y,x);
        }
        find_fa(1,-1);
        printf("Case #%d:\n",++cns);
        for(int i=0; i<q; i++)
        {
            scanf("%d",&m);
            for(int j=0; j<m; j++)
            {
                scanf("%d",&qn[j]);
                im[qn[j]]=0;
            }
            cnt=n-m;
            for(int j=0; j<m; j++)
            {
                if(im[qn[j]]==0)
                    solve(qn[j]);
            }
            for(int j=0; j<m; j++)
            {
                im[qn[j]]=1;
            }
            cout<<cnt<<endl;
        }
    }
}