PAT 1118 Birds in Forest

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该博客讨论了一种利用并查集（Disjoint Set）数据结构解决科学家在森林中拍摄鸟类照片后进行计数和判断同树问题的方法。通过输入图片中每张照片包含的鸟的编号，程序可以合并属于同一棵树的鸟，并最终确定森林中可能的最大树木数量。在处理查询时，程序能够快速判断任意两只鸟是否属于同一棵树。遇到的错误是起初忘记返回`findfather`函数的值，导致段错误，修复后程序能够正确运行。

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 (≤10
4
) which is the number of pictures. Then N lines follow, each describes a picture in the format:

K B
1
B
2
… B
K

where K is the number of birds in this picture, and B
i
'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 10
4
.

After the pictures there is a positive number Q (≤10
4
) 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.
在这里插入图片描述
题解：
在输入的时候用Union函数和findfather函数将同一幅画的鸟儿放在一个集合里。因为题中说过鸟儿的标号是个连续的序列，那么鸟儿的数量其实就是最大的标号。 因为输入的顺序不同可能会造成本来需要更新的father[i]未更新，所以我用了一个语句对其进行所有的更新，而如何判断有几个集合呢，我选择了把father[i]放入set中统计个数即可。
代码：

#include<iostream>
#include<set>
#include<vector>
using namespace std;
int father[10009];
set<int> group;
int findfather(int i){
	if(i==father[i]) return i;
	else{
		int a=i;
		while(i!=father[i]){
			i=father[i];
		}
		while(a!=father[a]){
			int z=a;
			a=father[a];
			father[a]=i;
		}
	}
    return i;
}
void Union(int i,int j){
	if(father[i]==father[j]) return;
	else{
		int z=findfather(i);
		father[findfather(j)]=z;
	}
}
int main(){
	int n,q,k;
	cin>>n;
	int max=0;
	for(int i=1;i<=10009;i++) father[i]=i;
	while(n--){
		cin>>k;
		vector<int> v(k);
		for(int i=0;i<k;i++){
			cin>>v[i];
			v[i]>max?max=v[i]:max=max;
		} 
		for(int i=1;i<k;i++) Union(v[0],v[i]);
	}
	for(int i=1;i<=max;i++) father[i]=findfather(i);
    for(int i=1;i<=max;i++) group.insert(father[i]);
	printf("%d %d\n",group.size(),max);
	cin>>q;
	while(q--){
		int i,j;
		cin>>i>>j;
		int fai,faj;
		fai=findfather(i);
		faj=findfather(j);
		if(fai==faj) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}