PTA甲级1118（压缩路径的并查集问题）

PTA甲级1118，压缩路径的并查集算法

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int leader[10005];
vector<int> trees;
int findleader(int bird){
	int temp=bird;
	//寻找最终leader 
    while(bird!=leader[bird]) 	
		bird=leader[bird];
	//压缩路径 
    while(temp!=leader[temp]){
        int t_temp = temp;
        temp = leader[temp];
        leader[t_temp] = bird;
    }
    return bird;
}

int main(){
	int n,q,num=0,tree=0;	cin>>n;
	for(int i=0;i<10005;i++)
		leader[i]=i;
	for(int i=0;i<n;i++){
		int k,lead;	cin>>k>>lead;
		num=max(num,lead);
		leader[lead]=findleader(lead);
		for(int j=1;j<k;j++){
			int curr;	cin>>curr;
			num=max(num,curr);
			//合并集合(仅将组Blead的leader[lead]改为组A的lead) 
			leader[findleader(curr)]=findleader(lead);
		}
	}
	//将组B所有成员的leader改为组A的lead 
	for(int i=1;i<=num;i++)
		findleader(i);
	for(int i=1;i<=num;i++)
		if(find(trees.begin(),trees.end(),leader[i])==trees.end()){
			trees.push_back(leader[i]);
			tree++;
		}
	cout<<tree<<" "<<num<<endl;
	cin>>q;
	for(int i=0;i<q;i++){
		int b1,b2;	cin>>b1>>b2;
		if(leader[b1]==leader[b2])	cout<<"Yes"<<endl;
		else	cout<<"No"<<endl;
	}
	return 0;
}