树的判定
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 }
