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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 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 Sample Output Case 1 is a tree. Case 2 is a tree. Case 3 is not a tree. 这个题总之挺坑的。 树的基本性质： ⒈必有一个特定的称为根(ROOT)的结点； 2.一棵树如果有n个节点，那么它一定恰好有n-1条边。 3.一棵树中的任意两个节点有且仅有唯一的一条路径连通。 4.在一棵树中加一条边将会构成回路； 对于这个题来说，不成立的条件有这几个： ①自己与自己形成自环。 ②这个节点加入的时候，与父节点已经有相同的节点了。 ③建树完成之后，有不止一棵树。 输入的时候有可能直接输入0 0，这个时候要输出它是一棵树，服了。。 代码如下：#include#include #include #include #include #include #define ll long longusing namespace std;const int maxx=1e5+100;struct node{ int x,y;}p[maxx];int a[maxx<<1];int f[maxx];inline int getf(int u){ if(u==f[u]) return f[u]; else return f[u]=getf(f[u]);}inline int merge(int u,int v){ int t1=getf(u); int t2=getf(v); if(t1==t2) return 1; f[t1]=t2; return 0;}int main(){ int x,y; int k=0; while(scanf("%d%d",&x,&y)) { if(x==-1&&y==-1) break; if(x==0&&y==0) { printf("Case %d is a tree.\n",++k); continue; } int cnt=0;int cnt1=0; p[++cnt].x=x,p[cnt].y=y; a[++cnt1]=x,a[++cnt1]=y; int flag=1; if(x==y) flag=0; while(scanf("%d%d",&x,&y),x||y) { p[++cnt].x=x,p[cnt].y=y; a[++cnt1]=x,a[++cnt1]=y; if(x==y) flag=0; } sort(a+1,a+1+cnt1); int len=unique(a+1,a+1+cnt1)-a-1; for(int i=0;i<=len;i++) f[i]=i; for(int i=1;i<=cnt;i++) { p[i].x=lower_bound(a+1,a+1+len,p[i].x)-a; p[i].y=lower_bound(a+1,a+1+len,p[i].y)-a; if(merge(p[i].x,p[i].y)) { flag=0; break; } } if(!flag) printf("Case %d is not a tree.\n",++k); else { for(int i=1;i<=len;i++) getf(i); int x=f[1]; for(int i=2;i<=len;i++) { if(f[i]!=x) { flag=0; break; } } if(!flag) printf("Case %d is not a tree.\n",++k); else printf("Case %d is a tree.\n",++k); } } return 0;}
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