How Many Tables HDU - 1213 （并查集）

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

2
5 3
1 2
2 3
4 5

5 1
2 5

Sample Output

2
4

 

译文

今天是伊格纳修斯的生日。他邀请了很多朋友。现在是晚餐时间。伊格纳修斯想知道他至少需要多少张桌子。你必须注意到并非所有的朋友都相互认识，并且所有的朋友都不想和陌生人呆在一起。 

这个问题的一个重要规则是，如果我告诉你A知道B，B知道C，那意味着A，B，C彼此了解，所以他们可以留在一个表中。 

例如：如果我告诉你A知道B，B知道C，D知道E，所以A，B，C可以留在一个表中，D，E必须留在另一个表中。所以Ignatius至少需要2张桌子。 

输入

输入以整数T（1 <= T <= 25）开始，表示测试用例的数量。然后是T测试案例。每个测试用例以两个整数N和M开始（1 <= N，M <= 1000）。N表示朋友的数量，朋友从1到N标记。然后M行跟随。每一行由两个整数A和B（A！= B）组成，这意味着朋友A和朋友B彼此了解。两个案例之间会有一个空白行。 

产量

对于每个测试用例，只输出Ignatius至少需要多少个表。不要打印任何空白。 

样本输入

2
5 3
1 2
2 3
4 5

5 1
2 5

样本输出

2
4

 

#include
#include
#include
#define MAX 1010
#define inf 0x3f3f3f3f
using namespace std;

int pre[MAX];
int n, m;

void init(int n){
	for(int i = 0; i <= n; i++){
		pre[i] = i;
	}
}

int find(int p){
	if(p == pre[p])	return p;
	else return pre[p] = find(pre[p]);
}

Union(int a,int b){
	int fa = find(a);
	int fb = find(b);
	if(fa != fb)
		pre[fb] = fa;
}

int main(){
	int T;
	cin>>T;
	while(T--){
		cin>>n>>m;
		init(n);
		int a,b;
		for(int i = 0; i < m; i++){
			cin>>a>>b;
			Union(a,b);
		}
		int ans = n;
		for(int i = 1; i <= n; i++){
			if(pre[i] != i){
				ans--;
			}
		}
		cout<