1118 Birds in Forest

最新推荐文章于 2022-10-02 14:45:00 发布

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本文介绍了一个利用并查集算法解决森林中鸟类计数问题的方法。通过分析图片中小鸟的位置来确定它们所属的树木数量及彼此间的关系。

1118 Birds in Forest
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 (≤10
4
) which is the number of pictures. Then N lines follow, each describes a picture in the format:

K B
1
B
2
… B
K

where K is the number of birds in this picture, and B
i
'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 10
4
.

After the pictures there is a positive number Q (≤10
4
) 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:
4
3 10 1 2
2 3 4
4 1 5 7 8
3 9 6 4
2
10 5
3 7
Sample Output:
2 10
Yes
No
分析：现在用相机给森林拍照，有小鸟在树上待着，在同一张照片中的小鸟则认为属于同一棵树。根据输入，问有多少棵树，以及总共有多少只鸟。
应用并查集的知识解决问题。
参考代码：

#include<iostream>
#include<cstring>
#define N 10008
using namespace std;
int father[N], maxn = 0;
int find(int x) {
	if (father[x] < 0)
		return x;
	else return father[x] = find(father[x]);
}
void Union(int a,int b){
	int rootA = find(a);
	int rootB = find(b);
	if (rootA == rootB)return;
	if (father[rootA] < father[rootB]) {
		father[rootA] += father[rootB];
		father[rootB] = rootA;
	}
	else {
		father[rootB] += father[rootA];
		father[rootA] = rootB;
	}
}
void check(int a, int b) {
	int rootA = find(a);
	int rootB = find(b);
	if (rootA == rootB)
		cout << "Yes" << endl;
	else cout << "No" << endl;
}
int main()
{
	memset(father, -1, sizeof(int)*N);
	int n, m, a, b, cnt = 0;
	scanf_s("%d", &n);

	int k;
	for (int i = 1; i <= n; i++) {
		scanf_s("%d", &k);
		int first, index;
		for (int j = 1; j <= k; j++) {
			scanf_s("%d", &index);
			if (index > maxn)maxn = index;
			if (j == 1)first = index;
			else Union(first-1, index-1);
		}
	}
	for (int i = 0; i < maxn; i++) 
		if (father[i] < 0)
			cnt++;
	cout << cnt << " " <<maxn<< endl;
	scanf_s("%d", &m);
	for (int i = 1; i <= m; i++) {
		scanf_s("%d%d", &a, &b);
		check(a - 1, b - 1);
	}

	return 0;
}

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