Birds in Forest

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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 10YesNo
这道题目很简单，直接用 Disjoint Set 就可以了，注意要路径优化，不然会超时。
#include <stdio.h>#include <stdlib.h>int a[10001]={0};int find(int i);int main(void){  int n;  scanf("%d",&n);  int N=n;  int max=0;  for(int j=0;j<N;j++){    int b;    scanf("%d",&b);      int c;    for(int k=0;k<b;k++){      if(k==0){               //将第一个元素储存，然后将后面的每个元素和它求并              scanf("%d",&c);        if(c>max)          max=c;      }      else{               int d;         scanf("%d",&d);         if(d>max)           max=d;         int x=find(d);         int y=find(c);         if(x!=y)             a[x]=y;      }    }  }  int h=0;  for(int i=1;i<max+1;i++){    if(a[i]<=0)      h++;  }  printf("%d %d\n",h,max);  int x;  scanf("%d",&x);  for(int i=0;i<x;i++){    int a1,a2;    scanf("%d %d",&a1,&a2);    if(find(a1)==find(a2))      printf("Yes\n");    else      printf("No\n");  }  return 0;}int find(int i){  if(a[i]<=0)    return i;  else    return a[i]=find(a[i]); //路径优化}

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