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.

InputThe 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.
OutputFor 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.

思路：（參考大神部落格，方向性注意）不能成環，不能成樹林，任何一個節點的入度不能大於1；

ps：入度：有多少條邊指向這個點，出度就是從這個點出發有多少條邊；

#include<stdio.h>
int pre[100020],flag[100020];
int find(int x)//不能路徑壓縮，遞迴找根結點； 
{	
	if(pre[x]==x)
		return x;
	else
		return find(pre[x]);
}
int main()
{
	int m,n,i,k,p=1;
	while(scanf("%d %d",&m,&n)!=EOF)
	{
		if(m<0&&n<0)
			break;
		int tot=0,ans=0;
		if(m+n==0)//m=0,n=0時條件； 
		{
			printf("Case %d is a tree.\n",p++);
			continue;
		}
		for(i=1;i<=100010;i++)
		{
			pre[i]=i;
			flag[i]=0;
		} 
		flag[n]=flag[m]=1;//標記m,n出現過； 
		pre[n]=m;
		//這道題和前篇部落格hdu小希的迷宮類似,但這道題有方向性,詳看第一幅圖的方向; 
		//由n指向m，而不能由m指向n,這樣查詢父親結點時就能查出入度大於1的節點了;
		//可以讓m,n顛倒位置看看測試 1 2 3 2 0 0,這組資料is not a tree； 
		while(scanf("%d %d",&m,&n) && (m+n))
		{
			flag[m]=flag[n]=1; 
			int fx=find(m);
			int fy=find(n);
			if(fx!=fy)
				pre[n]=m;//同上； 
			else ans=1;
		}
		if(ans)//有環； 
		{
			printf("Case %d is not a tree.\n",p++);
			continue;
		}
		for(i=1;i<=100010;i++)
		{
			if(flag[i] && pre[i]==i)
				tot++;
			if(tot>1) break;
		}
		if(tot>1)//入度大於1了或者成樹林了； 
			printf("Case %d is not a tree.\n",p++);
		else printf("Case %d is a tree.\n",p++);
	}
	return 0;
}