hdoj 2120 Ice_cream's world I

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 814    Accepted Submission(s): 478

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

 

思路：求环的个数。输入数据有多组。

 

 

代码：

 

#include<stdio.h>#include<stdlib.h>#include<algorithm>#include<queue>using namespace std;int per[1100];int n,m,f;int init(){for(int i=0;i<n;i++){per[i]=i;}}int find(int x){int r;r=x;while(r!=per[r]){r=per[r];}per[x]=r;return r;}int join(int x,int y){int fx=find(x);int fy=find(y);if(fx!=fy)per[fx]=fy;elsef++;}int main(){int a,b,i;while(scanf("%d%d",&n,&m)!=EOF){f=0;    init();    for(i=0;i<m;i++)    {    scanf("%d%d",&a,&b);    join(a,b);    }    printf("%d\n",f);}return 0;}