描述
传送门：我是传送门
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.
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.
输出
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).
样例
输入
6 8 5 3 5 2 6 4
5 6 0 08 1 7 3 6 2 8 9 7 5
7 4 7 8 7 6 0 03 8 6 8 6 4
5 3 5 6 5 2 0 0
-1 -1
输出
Case 1 is a tree.
Case 2 is a tree.
Case 3 is not a tree.
思路
跟HDU1272类似
写错find()坑了自己半个小时
注意坑点
- 0 0 也算一棵树
- 不要忽略每一行输出的最后那个’.’
代码
1 /* 2 * ============================================== 3 * 4 * Filename: N.cpp 5 * 6 * Link: http://poj.org/problem?id=1308 7 * 8 * Version: 1.0 9 * Created: 2018/09/24 02时59分21秒 10 * Revision: none 11 * Compiler: g++ 12 * 13 * Author: 杜宁元 (https://duny31030.top/), duny31030@126.com 14 * Organization: QLU_浪在ACM 15 * 16 * ============================================== 17 */ 18 #include<cstdio> 19 #include<iostream> 20 #include<cstring> 21 #include<algorithm> 22 using namespace std; 23 #define clr(a, x) memset(a, x, sizeof(a)) 24 #define rep(i,a,n) for(int i=a;i<=n;i++) 25 #define pre(i,a,n) for(int i=n;i>=a;i--) 26 #define ll long long 27 #define max3(a,b,c) fmax(a,fmax(b,c)) 28 #define ios ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0); 29 const double eps = 1e-6; 30 const int INF = 0x3f3f3f3f; 31 const int mod = 1e9 + 7; 32 const int N = 110; 33 int f[N]; 34 bool vis[N]; 35 int find(int x) 36 { 37 if(f[x] != x) 38 { 39 f[x] = find(f[x]); 40 } 41 return f[x]; 42 } 43 void init() 44 { 45 for(int i = 0;i < N;i++) 46 { 47 f[i] = i; 48 vis[i] = 0; 49 } 50 } 51 52 int x,y; 53 int main() 54 { 55 ios 56 #ifdef ONLINE_JUDGE 57 #else 58 freopen("in.txt","r",stdin); 59 // freopen("out.txt","w",stdout); 60 #endif 61 int ca = 1; 62 int flag = 0; 63 int tmp = -1; 64 init(); 65 while(scanf("%d %d",&x,&y) && (x != -1 && y != -1)) 66 { 67 if(x == 0 && y == 0) 68 { 69 // printf("tmp = %d !!! ",tmp); 70 if(tmp != -1) 71 tmp = find(tmp); 72 // printf("tmp = %d !!! ",tmp); 73 for(int i = 0;i < N;i++) 74 { 75 if(vis[i]) 76 { 77 if(tmp != find(i)) 78 { 79 // printf("%d ",i); 80 flag = 1; 81 break; 82 } 83 } 84 } 85 printf("Case %d ",ca++); 86 if(flag) 87 { 88 printf("is not a tree. "); 89 } 90 else 91 { 92 printf("is a tree. "); 93 } 94 flag = 0; 95 tmp = -1; 96 init(); 97 continue; 98 } 99 // printf("x = %d y = %d ",x,y); 100 if(tmp == -1) 101 tmp = x; 102 // printf("tmp = %d ",tmp); 103 int fx = find(x); 104 int fy = find(y); 105 // printf("fx = %d fy = %d ",fx,fy); 106 vis[x] = 1; 107 vis[y] = 1; 108 // printf("vis[%d] = %d vis[%d] = %d ",x,vis[x],y,vis[y]); 109 if(fx == fy) 110 { 111 flag = 1; 112 } 113 else 114 { 115 // printf("f[%d] = %d ",fy,fx); 116 f[fy] = fx; 117 } 118 } 119 120 fclose(stdin); 121 // fclose(stdout); 122 return 0; 123 }