hdoj1213How Many Tables

How Many Tables

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 18681 Accepted Submission(s): 9276

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
直接模板，求桌子数量即求树根的数量！ 代码： 
#include<stdio.h>
#define max 1500
int set[max];
int table,person,cognize;
struct line 
{
    int start;
    int end;
}num[max];
int find(int p)//查找根节点 
{
    int t;
    int child=p;
    while(p!=set[p])
    p=set[p];
    while(child!=p)
    {
        t=set[child];
        set[child]=p;
        child=t;
    }
    return p;
}
void merge(int x,int y)//合并 
{
    int fx=find(x);
    int fy=find(y);
    if(fx!=fy)
    set[fx]=fy;
}
int main()
{
    int i,j,k,t,u=0;
    scanf("%d",&t);
    while(t--)
    {
        k=0;
    scanf("%d%d",&person,&cognize);
        for(i=1;i<=person;i++)
         {
             set[i]=i;
         }
         for(i=1;i<=cognize;i++)
          {
              scanf("%d%d",&num[i].start ,&num[i].end );
          }
          for(i=1;i<=cognize;i++)
           {
               if(find(num[i].start )!=find(num[i].end ))//判断是否成环即是否认识的人相重复 
               merge(num[i].start ,num[i].end );
           }
           for(i=1;i<=person;i++)//判断树的数量即桌子的数量！ 
            {
                if(set[i]==i)
                k++;
            }
            printf("%d\n",k);
    }
    return 0;
}
原题目：hdoj1213How Many Tables