树的判定

最新推荐文章于 2025-06-19 09:30:00 发布

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最新推荐文章于 2025-06-19 09:30:00 发布

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原文链接：http://www.cnblogs.com/qijinbiao/archive/2011/10/17/2215448.html

本文提供了一种通过输入边的描述来判断给定的图是否构成一棵树的方法。使用了并查集来处理连通性和环的问题，并通过节点的入度来检查是否存在多个根节点。

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判断输入的是不是一棵树

输入

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.

The number of test cases will not more than 20,and the number of the node will not exceed 10000.
The inputs will be ended by a pair of -1.样例输入: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 00 03 8 6 8 0 0
-1 -1

样例输出:

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

Case 4 is a tree.

Case 5 is not a tree.

代码如下：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 const int N=10000+100;
 6 int in[N],vis[N];
 7 int p[N];
 8 int find(int x) {return x==p[x]? x: p[x]=find(p[x]);}
 9 int main()
10 {
11     int a,b,n=0,cs=1,i;
12     bool istree=1;
13     for(i=0;i<N;i++) p[i]=i;
14     memset(vis,0,sizeof(vis));
15     while(cin>>a>>b)
16     {
17         if(a==b&&a==-1) break;
18         if(a!=0&&b!=0)
19         {
20             n=max(n,max(a,b));
21             in[b]++;vis[a]=1;vis[b]=1;
22             if(istree&&in[b]>1)
23             {
24                 istree=0;printf("Case %d is not a tree.\n",cs++);
25             }
26             if(istree)
27             {
28                 int x=find(a),y=find(b);
29                 if(x!=y)
30                 p[y]=x;
31             }
32         }
33         else
34         {
35             if(istree)
36             {
37                 int t=0,s=0;
38                 for(i=0;i<=n;i++)
39                 {
40                     if(vis[i]&&i==find(i)) t++;
41                     if(t>1) break;
42                     if(vis[i]&&in[i]==0) s++;
43                     if(s>1) break;
44                 }
45                 if(i==n+1&&s==1)
46                 printf("Case %d is a tree.\n",cs++);
47                 else
48                 printf("Case %d is not a tree.\n",cs++);
49 
50             }
51             istree=1;
52             for(i=0;i<n;i++) p[i]=i;
53             memset(vis,0,sizeof(vis));
54             memset(in,0,sizeof(in));
55         }
56     }
57     return 0;
58 }