POJ 1308-Is It A Tree 并查集

题目来源: http://acm.pku.edu.cn/JudgeOnline/problem?id=1308

 

解题报告:

 

首先判断这些点是否属于同一个集合: 如果两个点之间有边相连,则它们属于同一个集合

 

接着,判断是否存在一个点入度为0,即根

 

再判断,除了根以外的点,是否入度都为1,

 

如果满足上述条件,则为树.

 

#include <iostream> using namespace std; int cnt=0; typedef struct _node { int key; int rank; int degree; _node* parent; }node; node **s; node* find(int a) { for(int i=0;i<cnt;i++) if(a==(s[i]->key)) return s[i]; return NULL; } void makeSet(int a) { if(find(a)==NULL) { s[cnt]=new node; s[cnt]->key=a; s[cnt]->parent=s[cnt]; s[cnt]->degree=0; s[cnt]->rank=0; cnt++; } } node* findSet(node* a) { if(a->parent!=a) a->parent=findSet(a->parent); return a->parent; } void link(node* x, node * y) { if(x->rank > y->rank) y->parent=x; else { x->parent=y; if(x->rank==y->rank) y->rank++; } } void _union(int a,int b) { node *s=find(b); s->degree++; link(findSet(find(a)),findSet(s)); } int main() { int caseNum=0; int a,b; s=new node*[10001]; while(1) { scanf("%d%d",&a,&b); if(a==-1 && b==-1) break; else if(a==0 && b==0) { int n1=0,n2=0,n3=0; bool judge=true; for(int i=0;i<cnt;i++) { if(s[i]->parent==s[i]) n1++;//集合个数 if(s[i]->degree==0) n2++;//入度为0的个数 if(s[i]->degree==1) n3++;//入度为1的个数 } if(cnt!=0 && (n1!=1 || n2!=1 || n3!=cnt-1)) judge=false; if(judge) cout << "Case " << ++caseNum << " is a tree." << endl; else cout << "Case " << ++caseNum << " is not a tree." << endl; cnt=0; } else { makeSet(a); makeSet(b); _union(a,b); } } delete [] s; return 0; }

 

附录:

Is It A Tree?

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 9159		Accepted: 3126

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.