PAT A1118. Birds in Forest (25)

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#并查集

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本文介绍了一道关于并查集的应用题目，通过分析鸟类图片数据来确定森林中树木的最大数量及判断不同鸟类是否位于同一棵树上。文章详细展示了并查集的实现过程，包括查找根节点、路径压缩和集合合并等关键操作。

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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

思路分析：

简单题，使用并查集即可。人数等于id号的最大值。

题解：

#include <cstdio>
const int MAX = 10010;
int n, k;
int father[MAX], root[MAX] = {0};
int numr = 0, numb = 0;

int findf(int x){
	int a = x;
	while(x != father[x]) x = father[x];
	while(a != father[a]){
		int z = a;
		a = father[a];
		father[z] = x;
	}
	return x;
}

void Union(int a, int b){
	int fa = findf(a);
	int fb = findf(b);
	if(fa != fb) father[fb] = fa;
}

int main(){
	scanf("%d", &n);
	for(int i = 1; i < MAX; i++) father[i] = i;
	while(n--){
		int id1, id2;
		scanf("%d %d", &k, &id1);
		if(id1 > numb) numb = id1;
		k--;
		while(k--){
			scanf("%d", &id2);
			if(id2 > numb) numb = id2;
			Union(id1, id2);
		}
	}
	for(int i = 1; i <= numb; i++){
		root[findf(i)]++;
	}
	for(int i = 1; i <= numb; i++){
		if(root[i]) numr++;
	}
	printf("%d %d\n", numr, numb);
	scanf("%d", &n);
	while(n--){
		int v1, v2;
		scanf("%d %d", &v1, &v2);
		if(findf(v1) == findf(v2)) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}