HDU 1325 并查集

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
 
    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.
",++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.
",++cas):printf("Case %d is not a tree.
",++cas);
51     }
52     return 0;
53 }
原文地址:https://www.cnblogs.com/zzqc/p/7608826.html