Ice_cream's world I

Ice_cream's world I

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 1   Accepted Submission(s) : 1

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

简单并差集，求环路数量；

代码：

 #include<stdio.h>
 int wall[];
 int find(int x){
     int r=x;
     while(r!=wall[r])r=wall[r];
     int i=x,j;
     while(i!=r)j=wall[i],wall[i]=r,i=j;
     return r;
 }
 int main(){
     int N,M,temp1,temp2,f1,f2,flot;
     while(~scanf("%d%d",&N,&M)){
         for(int i=;i<N;++i)wall[i]=i;
         flot=;
         while(M--){
             scanf("%d%d",&temp1,&temp2);
             f1=find(temp1);f2=find(temp2);
             if(f1!=f2)wall[f1]=f2;
             else flot++;
         }
         printf("%d\n",flot);
     }
     return ;
 }