Ice_cream's world I

最新推荐文章于 2020-04-06 20:28:26 发布

lkwan123

最新推荐文章于 2020-04-06 20:28:26 发布

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分类专栏：并查集

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本文介绍了一个基于并查集算法的土地分配问题解决方案。在一个由观察塔和围墙构成的地图中，通过并查集确定可以分配给勤奋ACMer的最大土地数量。文章提供了完整的C语言实现代码，并解释了如何通过跟踪节点的根来确定独立区域。

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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

Sample Output

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;
}