Species Tree uses HashTable to implement instance code

Given a species evolution map，
The relationship is expressed as follows：
x y : expressxforyThe ancestors。
A species will have only one ancestor，
An ancestor can have many evolved species，
Please find out the grandfather species of each question to ask about species(先祖The ancestors)，
每个物种会使用一个不重复的编号来express，
If the species does not have a grandfather species or does not exist，
Then please consider his grandfather's species0。(Born out of thin air)
Ensure that all species are connected，
And there will be no repeated evolution，
That is, the evolution map will only be a tree。

Input format:
Only one set of tests。
There will be two numbers on the first rowN、Q，Represents a total ofNSpecies andQA question。
NextN-1OK，每一OK有两个数字x、y，
The meaning is as stated in the title。
Next的QOK，每一OK有一个数字Z，
Represents the species number to be asked。
Measurement scope：
1 < N < 10000
0 < Q < 1000
0 < x, y, z < 1000000

Output format：
For each inquired species number，Add their grandfather species numbers and output them。

Enter sample：
6 3
1000 2000
1000 3000
2000 4000
3000 5000
5000 6000
5000
6000
1234

Output sample：
4000

Time limit：100msMemory limit：16000kb

#include <>
#include <>

int main(){
  int arrBucket[1000000];
  int N, Q;
  int x, y, z;
  long long sum = 0;

  memset(arrBucket, 0, sizeof(arrBucket));

  scanf("%d %d", &N, &Q);
  while(N -- > 1){
    scanf("%d %d", &x, &y);
    arrBucket[y] = x;
  }

  while(Q --){
    scanf("%d", &z);
    if(arrBucket[z] != 0){
      if(arrBucket[arrBucket[z]] != 0){
        sum += arrBucket[arrBucket[z]];
      }
    }
  }

  printf("%lld", sum);

  return 0;
}

#include <>
#include <>
#define NULLKEY -1

typedef struct {
  int *elem; //element  int *elemP; // Parent element  int count;
}HashTable;

int Hash(HashTable H, int k){
  return k % ;
}

void InitHashTable(HashTable *H){
  int i;

  H->elem = (int *)malloc(sizeof(int) * H->count);
  H->elemP = (int *)malloc(sizeof(int) * H->count);
  for(i = 0; i < H->count; i ++){
    H->elem[i] = NULLKEY;
    H->elemP[i] = NULLKEY; 
  }
}

void InsertHash(HashTable *H, int key, int value){
  int addr;

  addr = Hash(*H, key);
  while(H->elem[addr] != NULLKEY){
    addr = (addr + 1) % H->count;
  }

  H->elem[addr] = key;
  H->elemP[addr] = value;
}

int FindHash(HashTable *H, int key, int *addr){

  *addr = Hash(*H, key);

  while(H->elem[*addr] != key){
    *addr = (*addr + 1) % H->count;
    if(H->elem[*addr] == NULLKEY || *addr == Hash(*H, key)){
      return 0;
    }
  }

  return 1;
}

int main(){
  int N, Q;
  int x, y, z, addr;
  long long sum = 0;
  HashTable H;

  scanf("%d %d", &N, &Q);
   = N - 1;
  InitHashTable(&H);

  while(N -- > 1){
    scanf("%d %d", &x, &y);
    InsertHash(&H, y, x);
  }

  while(Q --){
    scanf("%d", &z);
    if(FindHash(&H, z, &addr)){
      if(FindHash(&H, [addr], &addr)){
        sum += [addr];
      }
    }
  }

  printf("%lld", sum);

  return 0;
}

#include <iostream>
#include <map>
using namespace std;

int main() {
  int n, q;
  cin >> n >> q;

  map<int,int> mp; 
  map<int,int>::iterator it;
  int x, y, z;
  for (int i=1; i<n; ++i) {  
    cin >> x >> y;
    (pair<int,int>(y,x));  
  }

  int sum = 0;
  for (int i=0; i<q; ++i) {
    cin >> z;
    it = (z);  
    if (it != ()) {
      it = (it->second);  
      if (it != ())
        sum += it->second;  
    }
  }

  cout << sum;
  return 0;
}