G - Ice_cream's world I （并查集）

点击打开链接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 partitionthe 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.InputIn 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 hasa wall(A and B are distinct). Terminate by end of file.OutputOutput 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

#include <stdio.h>#include <string.h>int per[1005],count=0;void init(int n){for(int i=0 ; i<n ; i++){per[i]=i;			//初始化 } }   int find(int x) { 	int r=x;				//找上级  	while(per[r]!=r) 	{  		r=per[r];} 	int i=x,j;while(i!=r)		//压缩 {j=per[i];per[i]=r;i=j; }  return r;  }     void join(int x , int y)  {  	int fx=find(x),fy=find(y);  	if(fx!=fy)  	 {  		per[fx]=fy;}else{++count;}  }  int main(){int m,n,i,a,b;while(~scanf("%d%d",&n,&m)){init(n);while(m--){scanf("%d%d",&a,&b);join(a,b);}printf("%d\n",count);}return 0;}

这有一个并查集详解，讲得很好点击打开链接