我找不出自己的代码哪里错了你知道吗

1118. Birds in Forest (25)

时间限制

150 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

Some scientists took pictures of thousands of birds in a forest. Assume that all the birds appear in the same picture belong to the same tree. You are supposed to help the scientists to count the maximum number of trees in the forest, and for any pair of birds, tell if they are on the same tree.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive number N (<= 10⁴) which is the number of pictures. Then N lines follow, each describes a picture in the format:
K B1 B2 ... BK
where K is the number of birds in this picture, and Bi's are the indices of birds. It is guaranteed that the birds in all the pictures are numbered continuously from 1 to some number that is no more than 10⁴.

After the pictures there is a positive number Q (<= 10⁴) which is the number of queries. Then Q lines follow, each contains the indices of two birds.

Output Specification:

For each test case, first output in a line the maximum possible number of trees and the number of birds. Then for each query, print in a line "Yes" if the two birds belong to the same tree, or "No" if not.

Sample Input:

43 10 1 22 3 44 1 5 7 83 9 6 4210 53 7

Sample Output:

2 10Yes
No
我自己写的代码，我自己和我同学找了半天也没找到错误在哪，只得了17分，感觉很蓝瘦。以后闲着没事在来找吧。
#include<bits/stdc++.h>#define MAXN 10001using namespace std;int N,K,Q;int pre[MAXN];int a[MAXN];int book[MAXN];set<int> st;int find(int n){  int r=n;  while(pre[r]!=r) r=pre[r];    int x=n,y;  while(pre[x]!=r)  {    y=pre[x];    pre[x]=r;    x=y;  }    return r;}void union_(int x,int y){  int a=find(x);  int b=find(y);  if(a!=b)  pre[a]=b;}int main(){  cin>>N;  int i,j,treenum=0;  for(i=1;i<=MAXN;i++)  pre[i]=i;  memset(book,0,sizeof(book));  st.clear();  for(i=0;i<N;i++)  {    cin>>K;    for(j=0;j<K;j++)    {      cin>>a[j];      st.insert(a[j]);    }        for(j=0;j<K;j++)      find(a[j]);        int flag=0,tag;    for(j=0;j<K;j++)    {      if(book[a[j]]==1) //已经在一棵树里了       {        flag=1;        tag=j;        break;      }    }    if(flag==0)    {      treenum++;      book[a[0]]=1;       for(j=1;j<K;j++)      {        book[a[j]]=1;        union_(a[0],a[j]);      }    }    else    {      for(j=0;j<K;j++)      {       book[a[j]]=1;       union_(a[j],a[tag]);      }     }  }    cout<<treenum<<' '<<st.size()<<endl;    cin>>Q;  int sx,sy;  while(Q--)  {    cin>>sx>>sy;    if(find(sx)==find(sy))  cout<<"Yes"<<endl;    else cout<<"No"<<endl;  }  return 0;}



下面贴的是别人的代码，其实我对于并查集现在来说有一个很大的认识，就是对于Union_这个操作尽量用父亲节点进行操作。

#include<cstdio>  #include<algorithm>  using namespace std;  const int maxn = 10000 + 10;  bool flag[maxn] = { false };  int N, K, B;  int father[maxn];  int findFather(int x)  {      if (x == father[x])          return x;      else      {          int F = findFather(father[x]);          father[x] = F;          return F;      }  }  int main()  {      for (int i = 0; i < maxn; i++)      {          father[i] = i;      }      scanf("%d", &N); int count = 0;      for (int i = 0; i < N; i++)      {          scanf("%d", &K);          int bf;          for (int j = 0; j < K; j++)          {              scanf("%d", &B);              if (!flag[B])              {                  count++;                  flag[B] = true;              }              if (j == 0)              {                  bf = findFather(B);              }              else if(findFather(B) != findFather(bf))              {                  father[findFather(B)] = findFather(bf);              }          }      }      int trees = 0;      for (int i = 1; i <= count; i++)      {          if (father[i] == i)          {              trees++;          }      }      printf("%d %d\n", trees, count);      int Q;      scanf("%d", &Q);      int a, b;      for (int i = 0; i < Q; i++)      {          scanf("%d %d", &a, &b);          if (findFather(a) == findFather(b))              printf("Yes\n");          else              printf("No\n");      }      return 0;  }