1118. Birds in Forest (25)

最新推荐文章于 2020-05-17 17:39:40 发布

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本文介绍了一个关于并查集算法的应用实例，通过处理鸟类图片数据来解决科学家们面临的树数量统计及鸟类位置判断的问题。文章提供了完整的代码实现，并对关键步骤进行了详细解释。

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

问题分析：简单的并查集，只是这里根节点要每次更新，使叶子节点与根节点直接相连，最终计算树的数量的时候只要看有几个根节点就可以了

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

const int maxn = 10100;
int fa[10100];
int isroot[10100];
int bird_num,tree_num;

int root(int x)
{
	int tmp = x;
	
	//找到最终根节点 
	while(x != fa[x])
		x = fa[x];
	
	//所有点连接到根节点 
	while(tmp != fa[tmp])
	{
		int index = tmp;
		
		tmp = fa[tmp];
		fa[index] = x;
	}	
	
	return x;
} 

void Union(int a,int b)
{
	int fA = root(a);
	int fB = root(b);
	
	if (fA != fB)
		fa[fA] = fB;
}

int main()
{
	int n,k,q;
	
	cin >> n;
	for(int i=0; i<=maxn; i++)
	{
		fa[i] = i;
		isroot[i] = -1;
	}
	
	bird_num = -1;
	tree_num = 0;
	for(int i=0; i<n; i++)
	{
		cin >> k;
		int b[11000];
		for(int j=0; j<k; j++)
		{
			cin >> b[j];
			if (bird_num < b[j])
				bird_num = b[j];
			Union(b[j],b[0]);
		}
	}

	//有多少个根节点就有多少棵树 
	for(int i=1; i<=bird_num; i++)
	{
		isroot[root(i)] = 1;
	}
	
	for(int i=1; i<=bird_num; i++)
	{
		if (isroot[i] == 1)
			tree_num++;
	}
	
	printf("%d %d\n",tree_num,bird_num);
	cin >> q;
	
	for(int i=0; i<q; i++)
	{
		int a,b;
		
		cin >> a >> b;
		if (fa[a] == fa[b])
			cout << "Yes" << endl;
		else
			cout << "No" << endl; 
	}
	
	return 0;
}

问题分析：简单的并查集，只是这里根节点要每次更新，使叶子节点与根节点直接相连，最终计算树的数量的时候只要看有几个根节点就可以了