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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 判斷給出的一個有向圖是不是一顆樹，每個結點的入度只能是1（除了根）； 用並查集維護這顆樹，註意判斷入度以及結點編號不一定連續，離散化一下就好。

 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 }

HDU 1325 並查集