How Many Tables
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18165 Accepted Submission(s):
8942
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.
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
提示：不用管他输入的格式！
这个还是查找树的个数！父节点等于本身的点是根节点！
1 #include<stdio.h> 2 #include<string.h> 3 #include<algorithm> 4 #define maxn 10000 5 using namespace std; 6 int a; 7 int per[maxn]; 8 void init() 9 { 10 int i; 11 for(i=1;i<=a;i++) 12 per[i]=i;//初始化数组 13 } 14 int find(int x)//查找根节点 15 { 16 int t=x; 17 while(t!=per[t]) 18 t=per[t]; 19 per[x]=t;//缩短路径 20 return t; 21 } 22 void join(int x,int y) 23 { 24 int fx=find(x); 25 int fy=find(y); 26 if(fx!=fy) 27 per[fx]=fy; 28 else 29 return ; 30 } 31 32 int main() 33 { 34 int b,n,c,d; 35 scanf("%d",&n); 36 while(n--) 37 { 38 39 int sum=0,i; 40 scanf("%d%d",&a,&b); 41 init(); 42 while(b--) 43 { 44 scanf("%d%d",&c,&d); 45 join(c,d); 46 } 47 for(i=1;i<=a;i++) 48 { 49 if(per[i]==i) 50 sum++;//查找根节点的个数51 } 52 printf("%d ",sum); 53 } 54 return 0; 55 }