HDU 5927 Auxiliary Set （树形）

Auxiliary Set

Time Limit: 9000/4500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2035    Accepted Submission(s): 602

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：

求出每次询问中重要的点的个数。给你不重要的点，这些点如果是两个重要点的LCA，那这个不重要的就变成重要的了。

POINT：

给你的x个点之外的所有点都是重要的。那么只要看不重要的点里有几个重要的。

遍历数，存下fa是谁和son的个数，和深度。

这x个点按深度排序。那么从最深的开始。

如果son>=2，那么就是重要的点，（即肯定有两个重要的点的LCA是他——仔细想想）。

son==1,这个点肯定不是，但是这个点的重要的son可以通过这个点来使其他点变重要。所以这个分支还需要，不能son--.  

son=0，那肯定不是，而且要把这个点的fa的son--。因为不需要这个分支了。

#include <stdio.h>#include<iostream>#include <math.h>#include<algorithm>#include <string.h>using namespace std;const int maxn = 2e5+10;int head[maxn],nxt[maxn],sz;int to[maxn];int fa[maxn],son[maxn],dep[maxn];int d;void dfs(int u,int pre){    fa[u]=pre;    dep[u]=d++;    for(int i=head[u];i!=-1;i=nxt[i]){        int v=to[i];        if(v==pre) continue;        dfs(v,u);        son[u]++;    }}bool cmd(int a,int b){    return dep[a]>dep[b];}int main(){    int T;    scanf("%d",&T);    int cas=1;    while(T--){        sz=0;        d=1;        int n,q;        scanf("%d %d",&n,&q);        fa[n]=son[n]=0;        for(int i=1;i<=n;i++) head[i]=-1;        for(int i=1;i<=n-1;i++){            fa[i]=son[i]=0;            int u,v;            scanf("%d %d",&u,&v);            to[sz]=v;            nxt[sz]=head[u];head[u]=sz++;            to[sz]=u;            nxt[sz]=head[v];head[v]=sz++;        }        dfs(1,0);        int temp[maxn];        int num[maxn];        printf("Case #%d:\n",cas++);        while(q--){            int x;scanf("%d",&x);            int ans=n-x;            for(int i=1;i<=x;i++) scanf("%d",&num[i]);            sort(num+1,num+1+x,cmd);            for(int i=1;i<=x;i++) temp[num[i]]=son[num[i]];            for(int i=1;i<=x;i++){                if(temp[num[i]]>1){                    ans++;                }else if(temp[num[i]]==0){                    temp[fa[num[i]]]--;                }            }            printf("%d\n",ans);        }    }    return 0;}