hdu 2120 (Ice_cream's world I)

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): 1280    Accepted Submission(s): 765


[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

题意： 问有多少环（查了才知道，反正我没读懂）

#include<cstdio>
using namespace std;
int n,sum;
int fa[1010];
void init(){
for (int i=0;i<n;i++){
fa[i]=i;
}
}
int find (int x){
if (x!=fa[x])
fa[x]=find(fa[x]);
return fa[x];
}
void Union(int a,int b){
a=find(a);
b=find(b);
if (a==b) sum++;//成环
else{
fa[a]=b;
}
}

int main(){
int m;
int a,b;
while (scanf ("%d %d",&n,&m)!=EOF){
init();
sum=0;
while (m--){
scanf ("%d %d",&a,&b);
Union(a,b);
}
printf ("%d\n",sum);
}
return 0;
}