杭电 2120 Ice_cream's world I （并查集求环数）

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

Sample Output



大意：有n个碉堡，m种关系每个关系表示两个碉堡连在一起，问一共有多少环。
父结点相等就有一个环。

 1 #include<cstdio>
 2 int n,m,fa[1005],i,a,b,sum;
 3 int find(int a)
 4 {
 5     int r=a;
 6     while(r != fa[r])
 7     {
 8         r=fa[r];
 9     }
10     int i=a,j;
11     while(i != r)
12     {
13         j=fa[i];
14         fa[i]=r;
15         i=j;
16     }
17     return r;
18 }
19 void f1(int x,int y)
20 {
21     int nx,ny;
22     nx=find(x);
23     ny=find(y);
24     if(nx != ny)
25     {
26         fa[nx]=ny;
27     }
28     else
29     {
30         sum++;
31     }
32 }
33 int main()
34 {
35     while(scanf("%d %d",&n,&m)!=EOF)
36     {
37         sum=0;
38         for(i = 0 ; i < n ; i++)
39         {
40             fa[i]=i;
41         }
42         for(i = 0 ; i < m ; i++)
43         {
44             scanf("%d %d",&a,&b);
45             f1(a,b);
46         }
47         printf("%d
",sum);
48     }
49 }

——将来的你会感谢现在努力的自己。