hdoj 2120 Ice_cream's world I 【判断成环】

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1092    Accepted Submission(s): 657

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
灯看做点，构图，判断成环个数【以前写的，补贴上】
#include<stdio.h>int per[1010];int find(int p){    int r=p;    int t;    while(p!=per[p])    {        p=per[p];     }      while(r!=p)     {         t=per[p];         per[p]=p;         r=t;     }     return p;}void merge(int x,int y){    int fx=find(x);    int fy=find(y);    if(fx!=fy)        per[fx]=fy;}int main(){    int n,m,a,b;    int ans,fa,fb;    while(~scanf("%d%d",&n,&m))    {        ans=0;        for(int i=0;i<n;i++)        {            per[i]=i;        }        for(int i=0;i<m;i++)        {            scanf("%d%d",&a,&b);            fa=find(a);            fb=find(b);            if(fa==fb)//判断是否成环                 ans++;            else                merge(a,b);        }        printf("%d\n",ans);     }    return 0;}