题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=1213
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
2
5 3
1 2
2 3
4 5
5 1
2 5
Sample Output
2
4
题意描述：
输入人数和关系数以及关系
计算并输出有多少个独立团体
解题思路：
处理数据，使用并查集算法即可。
AC代码：
1 #include<stdio.h>
2 void merge(int u,int v);
3 int getf(int v);
4 int f[1010];
5 int main()
6 {
7 int T,n,m,a,b,i,sum;
8 scanf("%d",&T);
9 while(T--)
10 {
11 scanf("%d%d",&n,&m);
12 for(i=1;i<=n;i++)
13 f[i]=i;
14 while(m--)
15 {
16 scanf("%d%d",&a,&b);
17 merge(a,b);
18 }
19
20 for(sum=0,i=1;i<=n;i++)
21 {
22 if(f[i]==i)
23 sum++;
24 }
25 printf("%d\n",sum);
26 }
27 return 0;
28 }
29 void merge(int u,int v)
30 {
31 int t1,t2;
32 t1=getf(u);
33 t2=getf(v);
34 if(t1 != t2)
35 f[t2]=t1;
36 }
37 int getf(int v)
38 {
39 if(f[v]==v)
40 return v;
41
42 return f[v]=getf(f[v]);
43 }