HDOJ 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): 974 Accepted Submission(s): 566



[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

嗯……虽然把题看懂了，但是，成环的代码是硬记下来的。所以刚上手的时候有种没有头绪的感觉。

但是，并查集，意外的好理解啊。

#include<stdio.h>
int per[1001];
int count;
int Find(int r)
{
while(r!=per[r])
r=per[r];
return r;
}
void Merge(int x,int y)
{
int fx=Find(x);
int fy=Find(y);
if(fx!=fy)
per[fx]=fy;
else
count++;
}
int main()
{
int n,m,a,b,i;
while(~scanf("%d%d",&n,&m))
{
count=0;
for(i=0;i<n;i++)
per[i]=i;
for(i=1;i<=m;i++)
{
scanf("%d%d",&a,&b);
Merge(a,b);
}
printf("%d\n",count);
}
return 0;
}