PAT (Advanced Level) Practice 1118 Birds in Forest（25 分）

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博客围绕科学家拍摄森林中鸟类照片的问题展开，已知同一张照片中的鸟属于同一棵树，需计算森林中树的最大数量，并判断任意两只鸟是否在同一棵树。解题思路采用并查集方法。

1118 Birds in Forest（25 分）

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:

Sample Output:

2 10
Yes
No

题意：出现在同一张照片的bird属于同一棵树，问一共有几棵树和询问的两只小鸟是否在同一棵树上。

思路：并查集。

代码：

#include <bits/stdc++.h>
using namespace std;
const int maxn=1e4+5;
int father[maxn],n;
void init()
{
    for(int i=0;i<=maxn;i++)father[i]=i;
}
int findroot(int x)
{
    if(father[x]==x)return x;
    return father[x]=findroot(father[x]);
}
void Union(int x,int y)
{
    int a=findroot(x);
    int b=findroot(y);
    if(a!=b)father[a]=b;
}
int main()
{
    int n;scanf("%d",&n);
    init();
    int mm=0,k,x,y,vis[maxn]={0};
    for(int i=0;i<n;i++)
    {
        scanf("%d",&k);
        scanf("%d",&x);
        mm=max(mm,x);
        vis[x]=1;
        for(int j=1;j<k;j++)
        {
            scanf("%d",&y);
            vis[y]=1;
            Union(x,y);
            mm=max(mm,y);
        }
    }
    int cnt=0,num=0;
    for(int i=1;i<=mm;i++)
    {
        if(vis[i]){
            if(findroot(i)==i)cnt++;
            num++;
        }
    }
    printf("%d %d\n",cnt,num);
    int q;scanf("%d",&q);
    while(q--)
    {
        scanf("%d%d",&x,&y);
        if(findroot(x)==findroot(y))printf("Yes\n");
        else printf("No\n");
    }
}