HDU-1213 How Many Tables

最新推荐文章于 2024-11-10 21:49:48 发布

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本文介绍了一种使用并查集算法解决朋友分组问题的方法。通过构建并查集数据结构，实现快速查找与合并操作，有效解决了如何最少分配餐桌数量的问题。文章详细解释了并查集的工作原理及其实现细节。

Problem Description

Today is Ignatius' birthday. He invites a lot of friends. Now it's dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want to stay with strangers.

One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table.

For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least.

Input

The input starts with an integer T(1<=T<=25) which indicate the number of test cases. Then T test cases follow. Each test case starts with two integers N and M(1<=N,M<=1000). N indicates the number of friends, the friends are marked from 1 to N. Then M lines follow. Each line consists of two integers A and B(A!=B), that means friend A and friend B know each other. There will be a blank line between two cases.

Output

For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.

并查集。如果是朋友，放到同一个集合里。求数有多少个集合，数有多少个代表元就行了。

#include<iostream>
#include<vector>
using namespace std;

struct node {
	int parent;
	int rank;
};

int Find(vector<node>& v, int x) {
	vector<int> path;
	while (v[x].parent != x) {
		path.push_back(x);
		x = v[x].parent;
	}
	for (int i = 0; i != path.size(); i++)
		v[path[i]].parent = x;
	return x;
}

void Union(vector<node>& v, int x, int y) {
	x = Find(v, x);
	y = Find(v, y);
	if (v[x].rank < v[y].rank) {
		v[x].parent = y;
	}
	else {
		v[y].parent = x;
		if (v[x].rank == v[y].rank)
			v[x].rank++;
	}
}

int main() {
	int case_number; cin >> case_number;
	for (int i = 0; i < case_number; i++) {
		int n, m; cin >> n >> m;
		vector<node> v(n);
		for (int i = 0; i < n; i++) {
			v[i].parent = i;
			v[i].rank = 1;
		}
		for (int j = 0; j < m; j++) {
			int n1, n2; cin >> n1 >> n2;
			Union(v, n1 - 1, n2 - 1);
		}
		vector<bool> isParent(n);
		for (int i = 0; i < n; i++) {
			isParent[Find(v, i)] = true;
		}
		int result = 0;
		for (int i = 0; i < n; i++)
			if (isParent[i])
				result++;
		cout << result << endl;
	}
}