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


[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

 

[align=left]Author[/align]
Wiskey
 

[align=left]Source[/align]
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;