HDU 5927 Auxiliary Set 【DFS+树】(2016CCPC东北地区大学生程序设计竞赛)

Auxiliary Set

Time Limit: 9000/4500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 873    Accepted Submission(s): 271

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

 

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题目链接：

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

题目大意：

　　T组数据(T<=1000)，对于每组数据，N(N<=100000)个点的一棵树，根节点为1，一个点在Set里需要满足下列情况之一：

　　　　1.这个点是特殊点　　2.这个点是两个特殊点的最近公共祖先(LCA)。

　　M(M<=100000)个询问，每次询问给一个Q(Q<=N)，表示N个点里面有Q个点不是特殊点，接下来是Q个点。求Set里有几个数。

　　∑Q<=100000，超过1000的N或∑Q的数据组数<=10

题目思路：

　　【DFS】

　　首先由于每组数据的树是一样的，先预处理出树的每个节点的深度d[x]，父亲fa[x]，度s[x](儿子个数，不算子孙)。

　　接下来对于每一个Q，将Q个数按照深度从深到浅排序，从最深的开始做，c[x]表示c的儿子子树全为非特殊点的个数。

　　如果当前节点的所有子孙都不是特殊点（c[x]=s[x]），则当前结点也不是特殊点，不需要加入Set，并且将x的父亲y=fa[x]的c[y]++

　　如果当前节点只有一颗子树含有特殊点，其他子树都不含有特殊点(s[x]-c[x]<=1)，则这个点不需要加入Set，但是y=fa[x]的c[y]不加（表示x这个子树中有特殊点）。

　　

//
//by coolxxx
//#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<iomanip>
#include<map>
#include<stack>
#include<queue>
#include<set>
#include<bitset>
#include<memory.h>
#include<time.h>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
//#include<stdbool.h>
#include<math.h>
#pragma comment(linker,"/STACK:1024000000,1024000000")
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define abs(a) ((a)>0?(a):(-(a)))
#define lowbit(a) (a&(-a))
#define sqr(a) ((a)*(a))
#define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
#define mem(a,b) memset(a,b,sizeof(a))
#define eps (1e-10)
#define J 10000
#define mod 1000000007
#define MAX 0x7f7f7f7f
#define PI 3.14159265358979323
#define N 100004
using namespace std;
typedef long long LL;
double anss;
LL aans;
int cas,cass;
int n,m,lll,ans;
int s[N],b[N],c[N],fa[N],last[N],d[N];
struct xxx
{
    int to,next;
}a[N+N];
void add(int x,int y)
{
    a[++lll].next=last[x];
    last[x]=lll;
    a[lll].to=y;
}
bool cmp(int a,int b)
{
    return d[a]>d[b];
}
void dfs(int now,int ffa)
{
    int i,to;
    s[now]=1;c[now]=0;d[now]=d[ffa]+1;fa[now]=ffa;
    for(i=last[now];i;i=a[i].next)
    {
        to=a[i].to;
        if(to==ffa)continue;
        dfs(to,now);
        s[now]++;
    }
}
void work()
{
    int i,x,mm;
    ans=n;
    scanf("%d",&mm);
    for(i=1;i<=mm;i++)scanf("%d",&b[i]);
    sort(b+1,b+1+mm,cmp);
    for(i=1;i<=mm;i++)
    {
        x=b[i];
        c[x]++;
        if(s[x]==c[x])c[fa[x]]++;
        if(s[x]-c[x]<2)ans--;
    }
    for(i=1;i<=mm;i++)c[b[i]]=c[fa[b[i]]]=0;
    printf("%d
",ans);
}
int main()
{
    #ifndef ONLINE_JUDGEW
//    freopen("1.txt","r",stdin);
//    freopen("2.txt","w",stdout);
    #endif
    int i,j,k;
    int x,y,z;
//    init();
//    for(scanf("%d",&cass);cass;cass--)
    for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
//    while(~scanf("%s",s))
//    while(~scanf("%d%d",&n,&m))
    {
        printf("Case #%d:
",cass);
        lll=0;mem(last,0);
        scanf("%d%d",&n,&m);
        for(i=1;i<n;i++)
        {
            scanf("%d%d",&x,&y);
            add(x,y),add(y,x);
        }
        dfs(1,0);
        for(i=1;i<=m;i++)
            work();
    }
    return 0;
}
/*
//

//
*/