HDU2120Ice_cream & amp; #39; s world I (basic and query set)

HDU2120Ice_cream & #39; s world I (basic and query set)
Ice_cream's world ITime Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 698 Accepted Submission (s): 398


Problem Descriptionice_cream's world is a rich country, it has alias 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 Wils 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 has a 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

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

Sample Output

3

AuthorWiskey: There are several loops in the figure.

#include
 
  const int N = 1015;int fath[N],n;void init(){    for(int i =0;i<=n;i++)        fath[i]=i;}int findfath(int x){    if(x==fath[x])        return fath[x];    fath[x]=findfath(fath[x]);    return fath[x];}int setfath(int x,int y){    x=findfath(x);    y=findfath(y);    if(x==y)        return 1;    fath[x]= y;    return 0;}int main(){    int x,y,m,ans;    while(scanf("%d%d",&n,&m)>0)    {        init();        ans=0;        while(m--)        {            scanf("%d%d",&x,&y);            ans+=setfath(x,y);        }        printf("%d\n",ans);    }}