Ice_cream's world I--hdu2120

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 839    Accepted Submission(s): 488

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.

 

 

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

 

注意：这是一个求总共有多少环（环内有节点的不算），我们可以利用判断两节点的根节点是否相同来判断！

 

 

 

 

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define maxn 10010
using namespace std;

int per[maxn],sum;
void init()
{
    int i;
    for(i=0;i<maxn;i++)
    {
        per[i]=i;//初始化数组
    }
}
int find(int x)//查找根节点
{
    int t=x;
    while(t!=per[t])
    t=per[t];
return t;
}
void join(int x,int y)
{
    int fx=find(x);
    int fy=find(y);
    if(fx==fy)
        sum++;//环的个数
    else
        per[fx]=fy;
}
int main()
{
    int a,b,i,m,n;
    while(scanf("%d%d",&a,&b)!=EOF)
    {
        init();
        sum=0;
        for(i=0;i<b;i++)
        {
            scanf("%d%d",&m,&n);
            join(m,n);
        }
        printf("%d
",sum);
    }
return 0;
}