HDU2120Ice_cream's world I（并查集+环）

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

题意：每一个用线围成的圈都可以赐给别人当领地，给你点与点的连线，问可以赐给多少人领地？

思路：并查集判断有多少个环就好了，判断环的方法在这题里有讲到。http://blog.csdn.net/zhaihao1996/article/details/52089036

代码：

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>#include<math.h>using namespace std;int pre[1010];int find(int x){int r=x;if(pre[r]!=r)pre[r]=find(pre[r]);return pre[r];}void join(int x,int y){int fx=find(x);int fy=find(y);if(fx!=fy)pre[fy]=fx;}int main(){int n,m;while(~scanf("%d%d",&n,&m)){int max=0;int b,c;for(int i=0;i<n;i++)pre[i]=i;for(int i=0;i<m;i++){scanf("%d%d",&b,&c);if(find(c)==find(b))max++;join(c,b);}printf("%d\n",max);}return 0;}