Ice_cream's world I--hdu2120

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 839    Accepted Submission(s):
488

Problem Description

ice_cream's world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partition the ice_cream’s world. But how many ACMers at most can be awarded by the queen is a big problem. One wall-surrounded land must be given to only one ACMer and no walls are crossed, if you can help the queen solve this problem, you will be get a land.

 

Input

In the case, first two integers N, M (N<=1000,
M<=10000) is represent the number of watchtower and the number of wall. The
watchtower numbered from 0 to N-1. Next following M lines, every line contain
two integers A, B mean between A and B has a wall(A and B are distinct).
Terminate by end of file.

 

Output

Output the maximum number of ACMers who will be
awarded.
One answer one line.

 

Sample Input

8 10

0 1

1 2

1 3

2 4

3 4

0 5

5 6

6 7

3 6

4 7

 

 

 

Sample Output

3

 

注意：这是一个求总共有多少环（环内有节点的不算），我们可以利用判断两节点的根节点是否相同来判断！

 

 

 

 

 #include<stdio.h>
 #include<string.h>
 #include<algorithm>
 #define maxn 10010
 using namespace std;

 int per[maxn],sum;
 void init()
 {
     int i;
     for(i=;i<maxn;i++)
     {
         per[i]=i;//初始化数组
     }
 }
 int find(int x)//查找根节点
 {
     int t=x;
     while(t!=per[t])
     t=per[t];
 return t;
 }
 void join(int x,int y)
 {
     int fx=find(x);
     int fy=find(y);
     if(fx==fy)
         sum++;//环的个数
     else
         per[fx]=fy;
 }
 int main()
 {
     int a,b,i,m,n;
     while(scanf("%d%d",&a,&b)!=EOF)
     {
         init();
         sum=;
         for(i=;i<b;i++)
         {
             scanf("%d%d",&m,&n);
             join(m,n);
         }
         printf("%d\n",sum);
     }
 return ;
 }