【题解】【PAT甲】1118 Birds in Forest (25 分)（并查集）

原创于 2021-12-17 16:39:13 发布 · 237 阅读

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本题解介绍了一道PAT甲级编程题的解决方法，利用并查集算法来确定图片中鸟类的最大分布数量及判断特定鸟类是否位于同一棵树上。通过使用map和数组等数据结构记录鸟类信息。

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题目链接

PTA | 程序设计类实验辅助教学平台

题目描述

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:

结尾无空行

Sample Output:

2 10
Yes
No

结尾无空行

题目大意

给你n棵张照片，每张照片上上有k只鸟，认为不同照片出现同一只鸟就说明在一棵树上，问最多能有几张照片拍的不是同一棵树，一共有几只鸟，然后任意给出两只鸟问再不在一棵树上

解题思路

并查集【算法】并查集_Elephant_King的博客-优快云博客

map用来记录一共有几只鸟

bird[i]用来记录编号为i的鸟在那张照片上

father代表照片之间的关系

isRoot代表最多有几张照片

标准模板可以参考

【题解】【PAT甲】1114 Family Property (25 分)（并查集）_Elephant_King的博客-优快云博客

【题解】【PAT甲】1107 Social Clusters (30 分)（并查集）_Elephant_King的博客-优快云博客

题解

#include<bits/stdc++.h>
using namespace std;
int father[10005];
int bird[10005];
int isRoot[10005];
map<int,int> ma;
int findFather(int x){
	int a=x;
	while(father[x]!=x)	x=father[x];
	while(father[a]!=a){
		int z=a;
		a=father[a];
		father[z]=x;
	}
	return x;
}
void Union(int a,int b){
	int fa=findFather(a);
	int fb=findFather(b);
	if(fa!=fb){
		father[fa]=fb;
	}
}
bool cmp(int a,int b){
	return a>b;
}
int main(){
	//freopen("input.txt","r",stdin);
	int n;
	cin>>n;
	for(int i=0;i<10005;i++)	father[i]=i;
	for(int i=1;i<=n;i++){
		int k;
		cin>>k;
		for(int j=0;j<k;j++){
			int temp;
			cin>>temp;
			ma[temp]=1;
			if(bird[temp]==0)	bird[temp]=i;
			Union(i,findFather(bird[temp]));
		}
	}
	int cnt=0;
	for(int i=1;i<=n;i++){
		isRoot[findFather(i)]++;
		if(isRoot[findFather(i)]==1)	cnt++;
	}
	sort(isRoot,isRoot+n+1,cmp);
	cout<<cnt<<" "<<ma.size()<<endl;
	int k;
	cin>>k;
	while(k--){
		int a,b;
		cin>>a>>b;
		if(findFather(bird[a])==findFather(bird[b]))	cout<<"Yes"<<endl;
		else	cout<<"No"<<endl;
	}
}