ACM Hong Kong online game b. Boxes

Original title URL: https://open.kattis.com/problems/boxes

Boxes

There is N boxes, indexed by a number from 1 to N. Each box may (or is not) is put into the other boxes. These boxes together form a tree structure (or a forest structure, to be precise).

Answer a series of queries of the following form:given a list of indices of the boxes, find the total number of boxes that the list of boxes actually contain.

Consider, for example, the following five boxes.

Figure 1: Sample input

• If the query is the list "1" and then the correct answer are "5", because box 1 contains all boxes.
• If the query is the list ' 4 5 ', then the correct answer is ' 2 ', for boxes 4 and 5 contain themselves and nothing else.
• If the query is the list "3 4" and then the correct answer is "2".
• If the query is the list "2 3 4" and then the correct answer is "4", since box 2 also contains box 5.
• If the query is the list "2", then the correct answer is "3", because box 2 contains itself and both other boxes.

Input

The first line contains the integer n (1≤n≤200000), the number of boxes.

The second line contains N

Integers. The ith integer is either the index of the box which contains the ith box, or zero if the ith b Ox isn't contained in any other box.

The third line contains an integer q(1≤q≤100000), the number of queries. The following Q lines'll has the following format:on each line, the first integer m (1≤m≤20 ) is the length of the list of boxes in this query, then M integers follow, representing the indices of the boxes .

Output

For each query, output a line which contains an integer representing the total number of boxes.

Sample Input 1	Sample Output 1
5 0 1 1) 2 2 5 1 1 2 4 5 2 3 4 3 2 3 4 1 2	5 2 2 4 3

Author (s): Chan Pak Hay

Source: Hong Kong Regional Online Preliminary 2016

Test instructions: N boxes, some of which are packed in other ones. Q Times asked, each time a given M-box, which asked this m-box contained altogether several boxes (including the M-box itself).

, if the first box is packed in box J, J is the parent node of I.

Box number 1~n, the box is a tree or a forest structure, determine a root node 0, so that the forest composed of boxes to synthesize a tree root node 0.

Starting from root node 0, Dfs traverses the entire tree, recording the sequential number of Dfs to each node dfn[i].

Tot[i] Indicates the total number of boxes contained in box I (including box I itself).

The collection of boxes contained in each box is sequential in the order of the DFS traversal.

Range[i] denotes the DFN number range of the box contained in box I, i.e. all chests DFN numbered Range[i].first~range[i].second are included in box I.

For example, a sample Build DFS Traversal :

Among them, range[1].first=1,range[1].second=5; DFN numbered chests are included in box 1.

range[2].first=2,range[2].second=4; The boxes numbered DFN are included in box 4.

range[3].first=5,range[3].second=5; The box with the DFN number 5 is included in Box 3.

range[4].first=3,range[4].second=3; The box with the DFN number 3 is included in box 4.

range[5].first=4,range[5].second=4; The box with the DFN number 4 is included in box 5.

Get the total number of boxes containing each box tot[i], and each box I contain the box number range Range[i],

For each set of queries Q, the violent enumeration once each box Q[i] is contained by another box, if Q[i] is included, delete Q[i], and finally the tot sum of the nodes not deleted.

#include <algorithm>#include<cstring>#include<string.h>#include<iostream>#include<list>#include<map>#include<Set>#include<stack>#include<string>#include<utility>#include<vector>#include<cstdio>#include<cmath>#defineLL Long Long#defineN 200005#defineINF 0X3FFFFFFusing namespacestd;intN;intBelong[n];//belong[i] means box I in box elong[i] insideintQnum;//Number of Queriesintm;intq[ -];vector<int>Vec[n];intDfn[n];//ordinal number of Dfs to each node Dfn[i]pair<int,int>range[N];//Range[i] Indicates the DFN number range of the box contained in box IintTot[n];//Tot[i] Indicates the total number of boxes contained in box I (including box I itselfintPos;voidDfsintU//DFS Traversal tree{Dfn[u]=pos++; Range[u].first=POS; Tot[u]=1;  for(intI=0; I<vec[u].size (); i++){        intv=Vec[u][i];        DFS (v); Tot[u]+=Tot[v]; } Range[u].second=POS; return;}intMain () { while(SCANF ("%d", &n)! =EOF) {POS=0; memset (Tot,0,sizeof(tot));  for(intI=1; i<=n;i++) vec[i].clear ();  for(intI=1; i<=n;i++) {scanf ("%d",&Belong[i]);    Vec[belong[i]].push_back (i); } DFS (0);//traversing from root node 0/*for (int i=1;i<=n;i++) {cout<<tot[i]<<endl;            cout<<dfn[i]<<endl;        cout<<range[i].first<< ' <<range[i].second<<endl; }  */scanf ("%d",&qnum);  while(qnum--) {scanf ("%d",&m);  for(intI=0; i<m;i++) {scanf ("%d", &q[i]);//q Array for the query of the box collection        }        intflag=0;//use flag to record which boxes have been deleted, and if (flag& (1<<i)), the box Q[i] is deleted.          for(intI=0; i<m;i++)//violence enumerate each box q[i] to see if it is contained in another box        {            if(! (flag& (1<<i )) { for(intj=i+1; j<m;j++)                    {                        if(! (flag& (1<<J )) {if(Range[q[i]].first<=range[q[j]].first && Range[q[i]].second>=range[q[j]].second) {//Box Q[j] included in box Q[i]Flag|= (1<<j); }                            Else if(Range[q[j]].first<=range[q[i]].first && Range[q[j]].second>=range[q[i]].second) {//Box Q[i] included in box Q[j]Flag|= (1<<i);  Break; }                        }                    }                }        }        intret=0;  for(intI=0; i<m;i++)        {            if(! (flag& (1<<i)) {//Box Q[i] not covered by other boxesret+=Tot[q[i]]; }} printf ("%d\n", ret); }   }    return 0;}

ACM Hong Kong online game b. Boxes