hdu 5927 Auxiliary Set（思路）

Auxiliary Set

Time Limit: 9000/4500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 813    Accepted Submission(s): 256

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 
 

详情看这篇题解，写的十分详细。

http://www.cnblogs.com/hahatianx/p/5943435.html

受益匪浅，还能从这个角度去做题...

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;#define N 100005struct Edge{    int next,v;} edge[N*4];struct Node{    int v,d;} node[N];int cnt,head[N];int son_num[N],dep[N],father[N],temp_son_num[N];void init(){    cnt=0;    memset(head,-1,sizeof(head));    memset(son_num,0,sizeof(son_num));}void addedge(int u,int v){    edge[cnt].v=v;    edge[cnt].next=head[u];    head[u]=cnt++;}int dfs(int u,int k,int d){    dep[u]=d;    father[u]=k;    for(int i=head[u]; i!=-1; i=edge[i].next)    {        int v=edge[i].v;        if(v==k) continue;        son_num[u]++;        dfs(v,u,d+1);    }}bool cmp(Node a,Node b){    return a.d>b.d;}int main(){    int T,n,m;    int Tcase=1;    int u,v,t,l;    scanf("%d",&T);    while(T--)    {        init();        scanf("%d %d",&n,&m);        for(int i=1; i<n; i++)        {            scanf("%d %d",&u,&v);            addedge(u,v);            addedge(v,u);        }        dfs(1,-1,1);        printf("Case #%d:\n",Tcase++);        while(m--)        {            scanf("%d",&t);            int ans=n-t,c=0;            for(int i=1; i<=t; i++)            {                scanf("%d",&l);                temp_son_num[l]=son_num[l];                node[c].v=l,node[c++].d=dep[l];            }            sort(node,node+c,cmp);            for(int i=0; i<c; i++)            {                int pos=node[i].v;                if(temp_son_num[pos]>=2) ans++;                else if(temp_son_num[pos]==0) temp_son_num[father[pos]]--;            }            printf("%d\n",ans);        }    }    return 0;}