【笨方法学PAT】1118 Birds in Forest （25 分）

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本文介绍了一种使用并查集数据结构来解决森林中树木计数及鸟类同树判断的问题。通过输入图片中鸟类的位置信息，算法能够高效地计算出森林中的树木数量，并判断任意两只鸟是否位于同一棵树上。文章详细阐述了并查集的初始化、查找、合并操作，并提供了完整的代码实现。

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一、题目

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

二、题目大意

一幅画里面的鸟为同一棵树上的，问有多少棵树和多少只鸟，以及对于两只鸟判断是否在同一个树上。

三、考点

并查集

四、注意

1、都是套路；

2、注意题目说：小鸟从1开始连续编号。

五、代码

#include<iostream>
#define N 10010
using namespace std;
int fa[N];

void init() {
	for (int i = 0; i < N; ++i)
		fa[i] = i;
}

int findFather(int x) {
	int a = x;
	while (x != fa[x])
		x = fa[x];
	while (a != fa[a])
	{
		int z = a;
		a = fa[a];
		fa[z] = x;
	}
	return x;
}
void Union(int a, int b) {
	int x = findFather(a);
	int y = findFather(b);
	if (x != y)
		fa[x] = y;
}

int main() {
	//init
	init();

	//read
	int n;
	cin >> n;
	int num_birds =0;
	for (int i = 0; i < n; ++i) {
		int m,a,b;
		cin >> m;
		for (int j = 0; j < m; ++j) {
			cin >> b;
			if (j == 0)
				a = b;
			else
				Union(a, b);
			//refresh num_birds
			if (b > num_birds)
				num_birds = b;
		}
	}

	//trees
	int num_trees = 0;
	for (int i = 1; i <= num_birds; ++i) {
		if (fa[i] == i) {
			num_trees++;
		}
	}

	//output the first line
	cout << num_trees << " " << num_birds << endl;

	//search
	cin >> n;
	for (int i = 0; i < n; ++i) {
		int a, b;
		cin >> a >> b;
		if (findFather(a) == findFather(b))
			cout << "Yes" << endl;
		else
			cout << "No" << endl;
	}

	system("pause");
	return 0;
}