HDOJ1213 How Many Tables(并查集)

最新推荐文章于 2017-08-20 21:53:22 发布

原创最新推荐文章于 2017-08-20 21:53:22 发布 · 957 阅读

1 ·

CC 4.0 BY-SA版权

HDOJ 同时被 2 个专栏收录

78 篇文章

订阅专栏

并查集

9 篇文章

订阅专栏

本文介绍了一个典型的并查集算法应用场景——解决朋友聚会时如何合理安排餐桌的问题。通过输入的朋友间的关系，使用并查集算法来计算最少需要准备多少张桌子以确保互相认识的人都能坐在一起。

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.

Sample Input

Sample Output

2
4

互相认识的人坐在一张桌子，问你需要准备多少张桌子。

并查集模板题。

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
int n, m, par[1010];
int find(int x)
{
	if(par[x] == x) return x;
	else return par[x] = find(par[x]);
}
void unite(int x, int y)
{
	x = find(x);
	y = find(y);
	if(x == y) return;
	par[y] = x;
}
void init()
{
	for(int i = 1; i <= n; ++i)
		par[i] = i;
}
int main(int argc, char const *argv[])
{
	int t;
	scanf("%d", &t);
	while(t--) {
		scanf("%d%d", &n, &m);
		init();
		for(int i = 0; i < m; ++i) {
			int a, b;
			scanf("%d%d", &a, &b);
			unite(a, b);
		}
		int ans = 0;
		for(int i = 1; i <= n; ++i)
			if(par[i] == i) ans++;
		printf("%d\n", ans);
	}
	return 0;
}