1118 Birds in Forest (25 point(s))

并查集算法解析

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本文深入探讨了并查集算法的应用，通过一个鸟类生态研究案例，详细解释了如何使用并查集来解决图中连通分支的问题。文章首先介绍了输入规格，包括图片数量、鸟类编号及查询对数。随后，通过代码示例展示了如何实现并查集算法，包括查找根节点、合并节点等关键操作。最后，输出了最大可能的树木数量和鸟类总数，以及对查询对的Yes/No判断。

1118 Birds in Forest (25 point(s))

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

考察点：并查集。即计算图中有多少个连通分支，给出两个点，判断两个结点是否在同一个连通分支。

1. 注意findRoot()；

2. 对于每一张照片的N只鸟，属于同一个连通分支，只需要N-1条边就需要把它们连接起来了。

#include<iostream>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAX = 1e4+7;
int Tree[MAX];
int a[MAX];
int findRoot(int x){
	if(Tree[x]==-1) return x;
	int temp = findRoot(Tree[x]);
	Tree[x]=temp;
	return temp;
}
int maxBird = 0;
int main(void){
	int N,k;cin>>N;
	memset(Tree,-1,sizeof(Tree));
	while(N--){
		cin>>k;
		for(int i=0;i<k;i++){
			cin>>a[i];
			maxBird = max(maxBird,a[i]);
		}
		for(int i=1;i<k;i++){
			int na = findRoot(a[i-1]);
			int nb = findRoot(a[i]);
			if(na!=nb) Tree[na]=nb;
		}
	}
	int cnt=0;
	for(int i=1;i<=maxBird;i++){
		if(Tree[i]==-1) cnt++;
	}
	cout<<cnt<<" "<<maxBird<<endl;
	int Q,x,y;cin>>Q;
	while(Q--){
		cin>>x>>y;
		x = findRoot(x);
		y = findRoot(y);
		if(x!=y) cout<<"No"<<endl;
		else cout<<"Yes"<<endl;
	}
	return 0;
}

AC时间：15min