并查集之POJ1308

Description

A tree is a well-known data structure that is either empty (null, void, nothing) or is a set of one or more nodes connected by directed edges between nodes satisfying the following properties.

There is exactly one node, called the root, to which no directed edges point.
Every node except the root has exactly one edge pointing to it.
There is a unique sequence of directed edges from the root to each node.
For example, consider the illustrations below, in which nodes are represented by circles and edges are represented by lines with arrowheads. The first two of these are trees, but the last is not.

In this problem you will be given several descriptions of collections of nodes connected by directed edges. For each of these you are to determine if the collection satisfies the definition of a tree or not.

Input

The input will consist of a sequence of descriptions (test cases) followed by a pair of negative integers. Each test case will consist of a sequence of edge descriptions followed by a pair of zeroes Each edge description will consist of a pair of integers; the first integer identifies the node from which the edge begins, and the second integer identifies the node to which the edge is directed. Node numbers will always be greater than zero.

Output

For each test case display the line "Case k is a tree." or the line "Case k is not a tree.", where k corresponds to the test case number (they are sequentially numbered starting with 1).

Sample Input

6 8  5 3  5 2  6 4
5 6  0 0

8 1  7 3  6 2  8 9  7 5
7 4  7 8  7 6  0 0

3 8  6 8  6 4
5 3  5 6  5 2  0 0
-1 -1

Sample Output

Case 1 is a tree.
Case 2 is a tree.
Case 3 is not a tree.

#include <stdio.h>

int i;
int s[1000];
int flag = 1;			//flag为0标志着不是树

void Initialize(int *s)		//初始化操作
{
	for (i = 1; i < 1000; i++)
		s[i] = i;
}

int Find(int x, int *s)		//查找操作,可以不用递归而用while实现
{
	if (s[x] == x)
		return x;
	else
		return Find(s[x], s);
}

void SetUnion(int n, int m, int *s)	//合并操作
{
	if (n == m)
		flag = 0;
	else if (s[m] == m)
		s[m] = n;
	else
		flag = 0;
}

int main()
{
	int n, m;
	int times = 1, first = 0, root;
	Initialize(s);
	while (scanf("%d %d", &n, &m) != EOF)
	{
		if (n < 0 || m < 0)
			break;
		if (n == 0 && m == 0)
		{
			for (i = 0; i < 1000; i++)
			{
				if (first == 0 && s[i] != i)	//找到第一个父节点不是本身的节点
				{
					first = 1;
					if (s[s[i]] == i)		//说明存在环
					{
						flag = 0;
						break;
					}
					root = Find(i, s);		//确定根节点
				}
				else if (first == 1 && s[i] != i)	
				{
					if (root != Find(i, s))	//看是否有一个根节点
						flag = 0;
				}
			}

			if (flag == 1)
				printf("Case %d is a tree.\n", times);
			else 
				printf("Case %d is not a tree.\n", times);
			times++;
			Initialize(s);
			flag = 1;
			first = 0;
		}
		else
		{
			SetUnion(n, m, s);		//执行合并，将m的父节点置为n
		}
	}
	return 0;
}