Auxiliary Set----DFS思维题

Auxiliary Set

Time Limit: 9000/4500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1083    Accepted Submission(s): 340

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

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5927

好一个思维DFS，这个题竟然能用dfs做我都没有想到，今天的比赛，我都没有看到这个题，放上大牛的链接，我从代码里讲吧。

http://blog.csdn.net/discreeter/article/details/52749361

代码：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int N=100010;struct node{    int to,next;}g[N*2];int cnt,head[N];int son[N],par[N],d[N],deep[N],tmp[N];bool vis[N];int n,m;void connect(int v,int u){//前向星存图    g[cnt].to=u;    g[cnt].next=head[v];    head[v]=cnt++;}void dfs(int v,int fa,int d){//一遍dfs求出父节点，儿子节点和深度    vis[v]=true;    par[v]=fa;    deep[v]=d;    for(int i=head[v];~i;i=g[i].next){        int u=g[i].to;        if(!vis[u])            dfs(u,v,d+1),son[v]++;    }}bool cmp(int a,int b){//按深度排序    return deep[a]>deep[b];}int main(){    int t;    scanf("%d",&t);    int ans=0;    while(t--){        ans++;        cnt=0;        int n,m;        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++){//初始化            vis[i]=false;            head[i]=-1;            son[i]=0;        }        for(int i=1;i<n;i++){            int u,v;            scanf("%d%d",&v,&u);            connect(u,v);            connect(v,u);        }        dfs(1,-1,1);        printf("Case #%d:\n",ans);        while(m--){            int k;            scanf("%d",&k);            int sum=n-k;//最开始有这么多重要节点            for(int i=1;i<=k;i++){                scanf("%d",&d[i]);                tmp[d[i]]=son[d[i]];//复制            }            sort(d+1,d+1+k,cmp);//按深度排序            for(int i=1;i<=k;i++){                if(tmp[d[i]]>=2)//大牛讲的很清楚这里                    sum++;                else if(tmp[d[i]]==0)//因为是从最深开始的，所有最深的那个不重要节点下面肯定全都是重要的节点                    tmp[par[d[i]]]--;            }            cout<<sum<<endl;        }    }    return 0;}