B

来源 hdu 1325

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.

和小希的迷宫有点像，但是他是有向的，要满足有向树的条件，那就是根节点没有被任何节点指向，其他的节点都只能被一个节点指向，否则就不对

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include <iomanip>
#include<cmath>
#include<float.h> 
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define mm(x,b) memset((x),(b),sizeof(x))
#include<vector>
#include<queue>
#include<stack>
#include<map>
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
typedef long long ll;
const ll mod=1e9+100;
const double eps=1e-8;
using namespace std;
const double pi=acos(-1.0);
const int inf=0xfffffff;
int pre[100005],temp,visit[100005],num[100005];//num是保存被指向的数目 
int find(int x)
{
	if(pre[x]==x) return x;
	return pre[x]=find(pre[x]);
}
void Union(int x,int y)
{
	int a=find(x),b=find(y);
	if(a==b)
	{
		temp=1;
		return ;
	}
	pre[b]=a;
}
int main()
{
	int x,y,ans=1;
	while(1)
	{
		temp=0;
		mm(visit,0);
		mm(num,0);
		int sum=0;
		rep(i,0,100005) pre[i]=i;
		while(1)
		{
			sf("%d%d",&x,&y);
			if(x+y==0) break;
			if(x+y<0) return 0;
			Union(x,y);
			visit[x]=1;visit[y]=1;
			num[y]++; 
		}
		rep(i,0,100005) 
		{
			if(visit[i]&&pre[i]==i) sum++;//判断是否是森林
			if(num[i]>1) temp=1;被指向超过两次
		}
		if(temp||sum>1)
		pf("Case %d is not a tree.
",ans++);
		else
		pf("Case %d is a tree.
",ans++);
	}
}