Appoint description:
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.
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
题目大意：
就是说，现在给你N个对象和T个关系，如果A和B有关系，B和C有关系，那么A和C也是有关系的。找出最后有几个这样不联通的集合.
解题思路：
直接裸裸的并查集，用并查集搞搞，直接出来，注意getf()和merge()函数的写法。
代码：
1 # include<cstdio> 2 # include<iostream> 3 4 using namespace std; 5 6 # define MAX 1004 7 8 int f[MAX]; 9 10 int getf ( int v ) 11 { 12 if ( f[v]==v ) 13 return v; 14 else 15 { 16 f[v] = getf(f[v]); 17 return f[v]; 18 } 19 } 20 21 22 void merge ( int v,int u ) 23 { 24 int t1 = getf(v); 25 int t2 = getf(u); 26 if ( t1!=t2 ) 27 { 28 f[t2] = t1; 29 } 30 } 31 32 int main(void) 33 { 34 int t;scanf("%d",&t); 35 while ( t-- ) 36 { 37 int n; 38 scanf("%d",&n); 39 for ( int i = 1;i <= n;i++ ) 40 { 41 f[i] = i; 42 } 43 int tt; 44 scanf("%d",&tt); 45 for ( int i = 1;i <= tt;i++ ) 46 { 47 int a,b; 48 scanf("%d %d",&a,&b); 49 merge(a,b); 50 51 } 52 int sum = 0; 53 for ( int i = 1;i <= n;i++ ) 54 { 55 if ( f[i]==i ) 56 sum++; 57 } 58 printf("%d ",sum); 59 60 } 61 return 0; 62 }