HDOJ 2120 Ice_cream's world I（并查集）

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 793    Accepted Submission(s): 464

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

8 100 11 21 32 43 40 55 66 73 64 7

 

Sample Output

3

 

大意：
两个塔之间有一个墙，就是求得把地图分成了几个区域，所以查看是否成环，如果成环，数量+1

ac代码：

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;#define MAXN 100100int pri[MAXN];int v[MAXN];int num;int find(int x){int r=x;while(r!=pri[r])   r=pri[r];int i=x,j;while(i!=r){j=pri[i];pri[i]=r;i=j;}return r;} void connect(int xx,int yy){int a=find(xx);int b=find(yy);if(a!=b)   pri[a]=b;else//如果成环，数量+1num++;}int main(){int n,m;int i,a,b;while(scanf("%d%d",&n,&m)!=EOF){for(i=0;i<n;i++)//因为有0号塔的存在，所以从0开始，不要习惯性从1开始pri[i]=i;num=0;for(i=0;i<m;i++){scanf("%d%d",&a,&b);connect(a,b);}printf("%d\n",num);}return 0;}