题目描述：
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
题意
lg邀请了许多朋友来吃饭。问需要几张桌子。 在同一张桌子上的人必须相互认识 。如果A认识B Ｂ认识Ｃ。 那么可以称Ａ与Ｃ相互认识。 这又是一个求并查集的题目。 我们可以发现，有多少个首领就需要多少张桌子。因为各个首领之间肯定是不认识的。
代码
1 #include<iostream>
2 #include<cstdio>
3 #include<algorithm>
4 using namespace std;
5 int fa[1005];
6 int findfa(int x){
7 return x==fa[x]?x:fa[x]=findfa(fa[x]);
8 }
9 void Union(int x,int y){
10 int fx=findfa(x);
11 int fy=findfa(y);
12 fa[fx]=fy;
13 }
14 int main(){
15 int t;
16 int n,k;
17 int x,y;
18 scanf("%d",&t);
19 while(t--){
20 scanf("%d %d",&n,&k);
21 for(int i=1;i<=n;i++){
22 fa[i]=i;
23 }
24 for(int i=0;i<k;i++){
25 scanf("%d %d",&x,&y);
26 Union(x,y);
27 }
28 int m=0;
29 for(int i=1;i<=n;i++){
30 if(fa[i]==i) m++;
31 }//m记录首领个数
32 printf("%d\n",m);
33 }
34 }