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.
输入描述
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.
输出描述
For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.
输入样例
2
5 3
1 2
2 3
4 5
5 1
2 5
输出样例
2
4
1 #include<iostream>
2 #include<cstring>
3 #define Maxn 1001
4 using namespace std;
5
6 int person[Maxn];//各自的上级
7
8 int FindBoss(int a)//找boss
9 {
10 if(person[a]==a)
11 {
12 return a;
13 }else{
14 return FindBoss(person[a]);
15 }
16 }
17
18 void FindFriend(int x,int y)
19 {
20 int xx,yy;
21 if(person[x]!=person[y])
22 {
23 xx=FindBoss(x);
24 yy=FindBoss(y);
25 person[xx]=yy;//合伙
26 }
27 }
28 int main()
29 {
30 int n,m,sum;//n:人数
31 int turn;
32 cin>>turn;
33 while(turn--)
34 {
35 memset(person,0,sizeof(int));//初始化
36 sum=0;
37 cin>>n>>m;
38 for(int i=1; i<=n; i++)
39 {
40 person[i]=i;//开始的时候他们的boss都是自己
41 }
42
43 int x,y;
44 for(int k=0; k<m; ++k)
45 {
46 cin>>x>>y;
47 FindFriend(x,y);
48 }
49 for(int i=1; i<=n; ++i)
50 {
51 if(person[i]==i)sum++;//只有boss的上级才是自己
52 }
53 cout<<sum<<endl;
54 }
55 return 0;
56 }
