Auxiliary Set

Time Limit: 9000/4500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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 uii 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

Source

2016CCPC东北地区大学生程序设计竞赛 - 重现赛

思路:记录儿子个数，子树的全删掉就是删掉一儿子；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=2e5+10,M=4e6+10,inf=1e9+10,mod=1e9+7;
const ll INF=1e18+10;
vector<int>v[N];
int si[N],fa[N],deep[N],a[N],change[N];
int n,m,q;
void init()
{
    for(int i=0;i<=n;i++)
    {
        change[i]=0;
        fa[i]=0;
        si[i]=0;
        deep[i]=0;
        v[i].clear();
    }
}
void dfs(int u,int pre,int dep)
{
    deep[u]=dep;
    for(int i=0;i<v[u].size();i++)
    {
        if(v[u][i]==pre)continue;
        si[u]++;dfs(v[u][i],u,dep+1);
        fa[v[u][i]]=u;
    }
}
int cmp(int a,int b)
{
    return deep[a]>deep[b];
}
int main()
{
    int T,cas=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&q);
        init();
        for(int i=1;i<n;i++)
        {
            int u,w;
            scanf("%d%d",&u,&w);
            v[u].push_back(w);
            v[w].push_back(u);
        }
        dfs(1,0,1);
        printf("Case #%d:\n",cas++);
        while(q--)
        {
            int ans=0;
            scanf("%d",&m);
            for(int i=1;i<=m;i++)
                scanf("%d",&a[i]);
            sort(a+1,a+1+m,cmp);
            for(int i=1;i<=m;i++)
            {
                int FA=fa[a[i]];
                if(si[a[i]]==0)
                {
                    si[FA]--;
                    change[FA]++;
                }
                if(si[a[i]]>=2)
                {
                    ans++;
                }
            }
            printf("%d\n",n-m+ans);
            for(int i=1;i<=m;i++)
            {
                int FA=fa[a[i]];
                if(change[FA])
                {
                    si[FA]+=change[FA];
                    change[FA]=0;
                }
            }
        }
    }
    return 0;
}