题目链接：https://vjudge.net/contest/271361#problem/E
具体思路：运用并查集，每一次连接上一个点，更新他的父亲节点，如果父亲节点相同，则构不成树，因为入读是2，然后再去遍历每一个点的父亲节点，判断一下祖宗节点有几个，只有1个才能构成树，注意0 0也是树.。。
AC代码：
1 #include<iostream>
2 #include<stack>
3 #include<queue>
4 #include<iomanip>
5 #include<stdio.h>
6 #include<algorithm>
7 #include<string>
8 #include<cstring>
9 #include<cmath>
10 #include<map>
11 using namespace std;
12 # define inf 0x3f3f3f3f
13 # define maxn 10000
14 int father[maxn];
15 int vis[maxn];
16 int k;
17 map<int,int>q;
18 int Find(int t)
19 {
20 return t==father[t]?t:father[t]=Find(father[t]);
21 }
22 void match(int t1,int t2)
23 {
24 int x=Find(t1);
25 int y=Find(t2);
26 if(x!=y)
27 {
28 father[y]=x;
29 return ;
30 }
31 else
32 k=1;
33 }
34 int main()
35 {
36 int x,y;
37 int num=0;
38 while(~scanf("%d %d",&x,&y))
39 {
40 q.clear();
41 k=0;
42 if(x==-1&&y==-1)break;
43 memset(vis,0,sizeof(vis));
44 if(x==0&&y==0)
45 {
46 printf("Case %d is a tree.
",++num);
47 continue;
48 }
49 for(int i=1; i<=maxn; i++)
50 {
51 father[i]=i;
52 }
53 vis[x]=1;
54 vis[y]=1;
55 match(x,y);
56 while(~scanf("%d%d",&x,&y))
57 {
58 if(x==0&&y==0)break;
59 match(x,y);
60 vis[x]=1;
61 vis[y]=1;
62 }
63 if(k==1)printf("Case %d is not a tree.
",++num);
64 else
65 {
66 int flag=0;
67 for(int i=1; i<=maxn; i++)
68 {
69 if(vis[i])
70 {
71 int temp=Find(i);
72 q[temp]=1;
73 }
74 }
75 if(q.size()==1) printf("Case %d is a tree.
",++num);
76 else printf("Case %d is not a tree.
",++num);
77 }
78 }
79 return 0;
80 }
81