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.
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.
记录一下刚A了这道题的我此时的心情。
重要的事情说三遍，，emmm 这道题巨坑！这道题巨坑！这道题巨坑！！！
平静一下
该题需要注意的是，空的图也是一棵树。
解题思路 ：1、判断是否有环。 2、判断入度为零的点只有一个。
判断是否有环：若边数大于等于顶点数，则有环。
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include<string> 5 #include<cmath> 6 #include<algorithm> 7 #include<queue> 8 #include<stack> 9 #include<deque> 10 #include<map> 11 #include<set> 12 #include<iostream> 13 using namespace std; 14 typedef long long LL; 15 const double pi=acos(-1.0); 16 const double e=exp(1); 17 const int N = 100000; 18 19 int captain[N],check[N]; 20 21 struct con{ 22 int x; 23 int y; 24 }con[N]; 25 26 int seek(int x) 27 { 28 // while(x!=captain[x]) 29 // x=captain[x]; 30 // return x; 31 return captain[x]; 32 } 33 34 void combine(int a,int b) 35 { 36 captain[b]=a; 37 // int boss1=seek(a); 38 // int boss2=seek(b); 39 // if(boss1!=boss2) 40 // captain[boss2]=boss1; 41 42 } 43 44 void init() 45 { 46 int i; 47 for(i=0;i<=N-1;i++) 48 captain[i]=i; 49 } 50 51 int main() 52 { 53 int i,p,j,n,a,b; 54 int cnt=1,flag=0,edge=0; 55 set<int > qq; 56 while(1) 57 { 58 qq.clear(); 59 flag=edge=0; 60 memset(con,0,sizeof(con)); 61 memset(captain,0,sizeof(captain)); 62 memset(check,0,sizeof(check)); 63 init(); 64 65 scanf("%d%d",&a,&b); 66 if(a==0&&b==0) 67 { 68 printf("Case %d is a tree. ",cnt); 69 cnt++; 70 continue; 71 } 72 if(a<0&&b<0) 73 break; 74 qq.insert(a); 75 qq.insert(b); 76 edge++; 77 con[0].x=a; 78 con[0].y=b; 79 combine(a,b); 80 p=1; 81 while(1) 82 { 83 scanf("%d%d",&a,&b); 84 if(a==0&&b==0) 85 break; 86 87 if(captain[a]==b||captain[b]==a) 88 flag=1; 89 90 qq.insert(a); 91 qq.insert(b); 92 edge++; 93 con[p].x=a; 94 con[p++].y=b; 95 96 combine(a,b); 97 } 98 int head=0; 99 for(i=0;i<p;i++) 100 { 101 if(captain[con[i].x]==con[i].x&&check[con[i].x]==0) 102 { 103 head++; 104 check[con[i].x]=1; 105 } 106 if(captain[con[i].y]==con[i].y&&check[con[i].y]==0) 107 { 108 head++; 109 check[con[i].y]=1; 110 } 111 if(con[i].x==con[i].y) 112 flag=1; 113 } 114 115 if(head!=1) 116 flag=1; 117 118 int tail=qq.size(); 119 if(edge>=tail) 120 flag=1; 121 122 if(flag==1) 123 printf("Case %d is not a tree. ",cnt); 124 else 125 printf("Case %d is a tree. ",cnt); 126 cnt++; 127 } 128 return 0; 129 }