hdu 5927 ( 思维 + 树 )

Auxiliary Set

Time Limit: 9000/4500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2254    Accepted Submission(s): 652

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

16 36 42 55 41 55 33 1 2 31 53 3 1 4

 

Sample Output

Case #1:363

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

题意：  给你一颗树 ，  然后每次询问给你c 个点为不重要的点，然后你要做的就是求出某个特定集合的 点的个数。 一个点若想在这个集合内，那么他就要满足两个条件中的一个，  第一 是重要点  二  是两个非重要的点的最近公共祖先。

思路 ：  我们先dfs 一遍建出一棵树 记录每个节点的儿子个数。   然后对于每次询问，我们需要对不重要的点进行单独建树，然后遍历每一棵树，求出树上满足是两非重要点的个数。  

代码： 

#include<bits/stdc++.h>
#define N 100005

using namespace std;

int head[N];
struct node
{
	int v;
	int next;
}edge[2*N];

int tot;
map<int ,int >mp;
vector<int >ve[N];
int n,q;
int num[N];
int sum[N];
int fa[N];
int ans;

void add(int u,int v){
	edge[++tot].v=v;
	edge[tot].next=head[u];
	head[u]=tot;
	return ;
}

void dfs(int u,int f){
	fa[u]=f;
	for(int i=head[u];i!=-1;i=edge[i].next){
		int v=edge[i].v;
		if(v==f) continue;
		dfs(v,u);
		sum[u]++;
	}
	return ;
}

int dfs1(int u,int f)
{
	if(mp[u]!=-1) return mp[u]; 
	
	int ssum=0;
	int son=0;
	int sz=ve[u].size();
	for(int i=0;i<sz;i++){
		int v=ve[u][i];
		if(v==f) continue;
		
		if(dfs1(v,u)>=1) ssum++;
		son++;
	}
	
	if(sum[u]-son+ssum>=2){
		ans++;
		return mp[u]=2;
	}
	if(sum[u]-son+ssum>=1){
		return mp[u]=1;
	}
	return mp[u]=-2;
}

int main()
{
	int T;
	int u,v;
	int kk=0;
	scanf("%d",&T);
	while(T--){
		memset(head,-1,sizeof(head));
		memset(sum,0,sizeof(sum));
		tot=0;
		
		scanf("%d %d",&n,&q);
		
		for(int i=1;i<n;i++){
			scanf("%d %d",&u,&v);
			add(u,v);
			add(v,u);
		}
		
		dfs(1,-1);
		/*
		for(int i=1;i<=n;i++) printf("%d ",sum[i]);
		printf("\n"); 
		*/
		
		
		printf("Case #%d:\n",++kk);
		while(q--){
			int cc;
			scanf("%d",&cc);
			mp.clear();
			for(int i=1;i<=cc;i++){
				scanf("%d",&num[i]);
				mp[num[i]]=-1;
				ve[num[i]].clear();
			}
			
			for(int i=1;i<=cc;i++){
				int f=fa[num[i]];
				if(mp[f]==-1){
					ve[f].push_back(num[i]);
				}
			}
			ans=0;
			for(int i=1;i<=cc;i++){
				if(mp[num[i]]==-1){
					dfs1(num[i],num[i]);
				}
			}
			
			ans=n-cc+ans;
			printf("%d\n",ans);
		}
		
	}
	return 0;
}