HD 2120 Ice_cream's world I 【并查集】

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 894 Accepted Submission(s): 522



[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]

3AC代码[code]#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int pre[1001],n;
void init()
{
for(int i=0;i<n;i++)
pre[i]=i;
}
int find(int x)//必须要用压缩路径，不用压缩路径过不了。-.-
{
if(pre[x]!=x)
{
pre[x]=find(pre[x]);
}
return pre[x];
}
int join(int x,int y)
{
int fx=find(x);
int fy=find(y);
if(fx==fy)
{
return true;
}
else
{
pre[fx]=fy;
return false;
}
}
int main()
{
int m,a,b,ans;
while(scanf("%d %d",&n,&m)!=EOF)
{
ans=0;
init();
for(int i=0;i<m;i++)
{
scanf("%d %d",&a,&b);

if(join(a,b))
{
ans++;
}
}
printf("%d\n",ans);
}
}

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