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.

判斷一個圖是否是一個有向樹，

１，全圖只有一個連通分量。

２，只有一個節點入度為零。

３．其餘節點入度必須為１。

判斷連通分量用並查集，判斷節點入度用陣列記錄每個節點的入度

#include <iostream>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 100100
using namespace std;
int n,m,flag,cas;
int Father[N];
int Rank[N];
int ind[N];
void UFDset()
{
    for(int i=1;i<N;i++){
        Father[i]=i;
        Rank[i]=0;
        ind[i]=0;
    }
}

int Find(int x)
{
    if(x!=Father[x])
        Father[x]=Find(Father[x]);
    return Father[x];
}

void Union(int R1,int R2)
{
    int r1=Find(R1),r2=Find(R2);
    if(Rank[r2]>Rank[r1]){
        Father[r1]=r2;
        Rank[r1]++;
    }
    else{
        Father[r2]=r1;
        Rank[r2]++;
    }
}

int main()
{
    UFDset();
    set<int>s;
    set<int>::iterator it;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n<0&&m<0)
            break;
        if(n==0&&m==0)
        {
            int cnt=0,rootnum=0;
            for(it=s.begin();it!=s.end();it++){
                if(Father[*it]==*it)
                    cnt++;
                if(ind[*it]==0)
                    rootnum++;
                if(ind[*it]>1)
                    flag=1;
            }
            if(rootnum!=1&&s.size()!=0)
                flag=1;
            if(cnt>1)
                flag=1;
            if(flag)
                printf("Case %d is not a tree.\n",++cas);
            else
                printf("Case %d is a tree.\n",++cas);
            UFDset();
            flag=0;
            s.clear();
            continue;
        }
        s.insert(n);
        s.insert(m);
        Union(n,m);
        ind[m]++;
    }
    return 0;
}