1118. Birds in Forest (25) <并查集+set>

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:

4
3 10 1 2
2 3 4
4 1 5 7 8
3 9 6 4
2
10 5
3 7

Sample Output:

2 10
Yes
No

就是求连通图的个数和是否在一个连通图上 典型的并查集

根据第一个出现的数字建立并查集

出现的数字储存在set里面，遍历set找到连通图的数量

最后判断find是否是一个树

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
int father[10000]; 
int find(int x){
	if(x==father[x]) return x;
	return father[x]=find(father[x]);
}
int main(){
	int n;
	cin>>n;
	set<int> s;
	for(int i=0;i<10000;i++) father[i]=i;
	for(int i=0;i<n;i++){
		int len,num;
		cin>>len;
        cin>>num;
        s.insert(num);
        num=find(num);
		for(int j=1;j<len;j++){
			int fz;
			cin>>fz;
			s.insert(fz);
			fz=find(fz);
			father[fz]=num;
		} 
	} 
	set<int> sum;
	for(set<int>::iterator it=s.begin();it!=s.end();it++){
		sum.insert(find(*it));
	} 
	cout<<sum.size()<<" "<<s.size()<<endl;
	int k;
	cin>>k; 
	for(int i=0;i<k;i++){
		int a,b;
		cin>>a>>b;
		if(find(a)==find(b)) cout<<"Yes"<<endl;
		else cout<<"No"<<endl;
	} 
	return 0;
}