1118. Birds in Forest (25)

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本文介绍了一种用于科学统计森林中鸟类数量及其分布情况的算法。通过分析一系列包含鸟类的照片，该算法能够确定森林中最大可能的树木数量及每棵树上的鸟类。此外，还能判断任意两只鸟是否位于同一棵树上。

1118. Birds in Forest (25)

时间限制

150 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <iostream>
#include <cstring>

using namespace std;

typedef long long int ll;
int a[10005];
int bird[10005];

int find_root(int *disj,int x){
	if(disj[x]<=0)
		return x;
	return disj[x] = find_root(disj,disj[x]);
}

void Union(int *disj,int root1,int root2){
	int num = disj[root1]+disj[root2];
	if( disj[root1]<disj[root2] ){
		if(disj[root2]==0)
			disj[root1] = num-1;
		else
			disj[root1] = num;
		disj[root2] = root1;
	}
	else{
		if(disj[root1]==0 && disj[root2]!=0)
			disj[root2] = num-1;
		else if(disj[root1]==0 && disj[root2]==0)
			disj[root2] = num-2;
		else
			disj[root2] = num;
		disj[root1] = root2;
	}
}


int main(){
	
	int N,K;
	scanf("%d",&N);
	int root1,root2;
	int num;
	for(int i=1; i<=N; i++){
		scanf("%d",&K);
		for(int i=1; i<=K; i++)
			scanf("%d",&a[i]);
		if(K==1){
			root1 = find_root(bird,a[1]);
			if(bird[root1]==0)
				bird[root1] = -1;
		}	
		for(int j=1; j<K; j++){
			root1 = find_root(bird,a[j]);
			root2 = find_root(bird,a[j+1]);
			if(root1!=root2)
				Union(bird,root1,root2);
		}	
	}
	int count=0,sum=0;
	for(int i=1; i<=10000; i++)
		if(bird[i]<0){
			count++;
			sum += -bird[i];
		}
	cout << count << " " << sum << endl;
	int Q;
	scanf("%d",&Q);
	int l,r;
	for(int i=1; i<=Q; i++){
		scanf("%d %d",&l,&r);
		root1 = find_root(bird,l);
		root2 = find_root(bird,r);
		if(root1==root2)
			cout << "Yes" << endl;
		else
			cout << "No" << endl;
	}
	return 0;
}