http://acm.hdu.edu.cn/showproblem.php?pid=1325

Is It A Tree?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26387    Accepted Submission(s): 6039

Problem Description

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.

Source

North Central North America 1997

判断给出的一个有向图是不是一颗树，每个结点的入度只能是1（除了根）；

用并查集维护这颗树，注意判断入度以及结点编号不一定连续，离散化一下就好。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<vector>
 6 #include<map>
 7 using namespace std;
 8 #define pii pair<int,int>
 9 #define inf 0x3f3f3f3f
10 int f[50005],tot[50005];
11 map<int,int>M;
12 int getf(int v){return f[v]==v?v:f[v]=getf(f[v]);}
13 int main()
14 {
15     int a,b,i,j,k;
16     int N,cas=0;
17     while(cin>>a>>b){bool ok=1;N=0;
18     if(a<=-1&&b<=-1)break;
19     M.clear();
20     if(a==0&&b==0){printf("Case %d is a tree.\n",++cas);continue;}
21     M[a]=++N;
22     if(a!=b)M[b]=++N;
23     else M[b]=1;
24         memset(tot,0,sizeof(tot));
25         for(i=1;i<=50005;++i)f[i]=i;
26         tot[M[b]]++;
27         int fa=getf(M[a]),fb=getf(M[b]);
28         if(fa!=fb) f[fa]=fb;
29         while(cin>>a>>b){if(a==0&&b==0) break;
30             if(!M[a])M[a]=++N;
31             if(!M[b])M[b]=++N;
32             a=M[a];
33             b=M[b];
34
35             fa=getf(a),fb=getf(b);
36             if(fa!=fb) f[fa]=fb;
37             N=max(N,max(a,b));
38             tot[b]++;
39         }
40         for(i=1;i<=N;++i)
41         {
42             if(tot[i]>1) {ok=0; break;}
43         }
44         int s=0;
45         for(i=1;i<=N;++i)
46         {
47             if(i==getf(i)) s++;
48             if(s>1){ok=0;break;}
49         }
50         ok?printf("Case %d is a tree.\n",++cas):printf("Case %d is not a tree.\n",++cas);
51     }
52     return 0;
53 }