杭电2120————简单并查集+判环

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 549    Accepted Submission(s): 308

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

 

题意很明确，就是说哨塔和墙围起来的那一块地可以被分给acmer，求这样的地最多有多少，其实就是求有多少个环。

这个题目我一开始...不会做..(说明对并查集理解的还是不够透彻）主要被难在了判环上。

后来经人点拨了一下明白了，比如说我们拿下面这个例子与函数find_root结合来看一下。

<span style="font-family:Microsoft YaHei;font-size:14px;">int find_root(int x)
{
	return ( x == set[x]) ? x : find_root(set[x]);
}</span>

这个函数的作用是找到点x所在树的根节点（初始化时根节点是自己)

假设已经构建好了如上图所示的图，然后我们读取一组数据 

3 4(表明3 4 边相连)

find_root(3)  =  0

find_root(4)  =  0

这两个有着共同的根节点，3,4又有关系，那么就相当于这样，于是就得到了一个环。

代码如下（亲测15MS）

<span style="font-family:Microsoft YaHei;font-size:14px;">#include <stdio.h>
#include <string.h>
#define maxn 1005

int ans,set[maxn];
void initialize()
{
	for(int i = 0 ; i < maxn ; i++)
		set[i] = i;
	ans = 0;
}
int find_root(int x)
{
	return ( x == set[x]) ? x : find_root(set[x]);
}
void Union(int x, int y)
{
	int root1 = find_root(x);
	int root2 = find_root(y);
	if(root1 != root2)
		set[root2] = root1;
}
int main()
{
	int N,M;
	int p1,p2;
	int root1,root2;
	while(scanf("%d%d",&N,&M) != EOF)
	{
		initialize();
		for(int i = 0 ; i < M ; i++)
		{
			scanf("%d%d",&p1,&p2);
			root1 = find_root(p1);
			root2 = find_root(p2);
			if(root1 == root2)
				ans ++ ;
			else
				Union(p1,p2);
		}
		printf("%d\n",ans);
	}
	return 0;
}</span>