hdu-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): 848    Accepted Submission(s): 494

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

 

Author

Wiskey

 

Source

HDU 2007-10 Programming Contest_WarmUp

 

# include<stdio.h># include<string.h># include<algorithm>using namespace std;int per[1020];int find(int x){    int j,k,r,i;    i = x, r = x;    while(r != per[r]) r=per[r];    while(i != r)    {        j=per[i];        per[j]=r;        i=j;    }    return r;}int main(){    int m,n;    while(~scanf("%d%d",&n,&m))    {        int i,t,s,a,b;        for(t=0;t<=n;t++)          per[t]=t;        for(i=0,s=0;i<m;i++)         {             scanf("%d%d",&a,&b);             int f1=find(a);             int f2=find(b);             if(f1==f2) s++;             else per[f1]=f2;         }         printf("%d\n",s);    }    return 0;