Ice_cream's world I
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 741 Accepted Submission(s):
429
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.
One answer one line.
Sample Input
8 10
0 1
1 2
1 3
2 4
3 4
0 5
5 6
6 7
3 6
4 7
Sample Output
3
#include<stdio.h>
#include<string.h>
#define MAX 1010
int set[MAX];
int sum=0;
int find(int fa)
{
int t;
int ch=fa;
while(fa!=set[fa])
fa=set[fa];
while(ch!=fa)
{
t=set[ch];
set[ch]=fa;
ch=t;
}
return fa;
}
void mix(int x,int y)
{
int fx,fy;
fx=find(x);
fy=find(y);
if(fx!=fy)
set[fx]=fy;
else
sum++;
}
int main()
{
int n,m,i,a,b;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(i=0;i<=n;i++)
set[i]=i;
for(i=0;i<m;i++)
{
scanf("%d%d",&a,&b);
mix(a,b);
}
printf("%d
",sum);
sum=0;
}
return 0;
}