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): 634 Accepted Submission(s): 368

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
Output the maximum number of ACMers who will be awarded.

One answer one line.

[align=left]Sample Input[/align]

8 10
0 1
1 2
1 3
2 4
3 4
0 5
5 6
6 7
3 6
4 7

[align=left]Sample Output[/align]

3

#include<iostream>
using namespace std;
int a[1005];
int sum;
int find(int r)
{
while(a[r]!=r)
r=a[r];
return r;
}
void merge(int x,int y)
{
int fx,fy;
fx=find(x);
fy=find(y);
if(fx!=fy)
a[fx]=fy;
else sum++;
}
int main()
{
int n,m,i;
while(cin>>n>>m)
{
int b,c;
sum=0;
for(i=0;i<n;i++)
a[i]=i;
for(i=0;i<m;i++)
{
cin>>b>>c;
merge(b,c);
}
cout<<sum<<endl;
}
return 0;
}