Ice_cream's world I

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

Author

Wiskey

Source

本题考查并查集，主要该思考的点式要围成一块地方，那么这个集合是联通的所以有相同根的时候就是一个地方，每相等一次，就加一次，以下是我写的代码。

#include<stdio.h>

int a[1000],n,m,d,b;

void fu()//给数组初始化，令值等于本身的角标

{

int i;

for(i=0;i<=n-1;i++)

a[i]=i; 

}

int fine(int x)//寻找根节点的函数

{

 int r=x;

 while(r!=a[r])

 r=a[r];

 int i,j;

 i=x;

 while(i!=r)

 {

 j=a[i];

 a[i]=r;

 i=j;

 }

 return r;

}

void join(int x,int y)//并两个集合的函数

{int fx,fy;

fx=fine(x);

fy=fine(y);

if(fx!=fy)

a[fx]=fy; 

}

int main()

{

 while(scanf("%d%d",&n,&m)!=EOF)

 {

 int c=0;

  fu();

  while(m--)

  {

   scanf("%d%d",&d,&b);

   if(fine(d)==fine(b))//每两个元素的根节点相同，就加一个地方

   c++;

   else

   join(d,b);

   

  }

  printf("%d\n",c);

 }

 

 return 0;

}