Is It A Tree?

链接：https://www.nowcoder.com/questionTerminal/1c5fd2e69e534cdcaba17083a5c56335
来源：牛客网
 

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.

 

输入描述:

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 and less than 10000.

 

输出描述:

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).

示例1

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 0
-1 -1

输出

Case 1 is a tree.
Case 2 is a tree.
Case 3 is not a tree.

题目解析：本题主要的要求是给出树的有向边的集合，然后让我们判断是否构成一棵树，判断是否构成一课树的定义为，所有节点中只有一个节点的入读为零，其余节点的入度都为一，因此我们需要存储树中所有边的集合，以及所有顶点的入度，然后去判断，在起始节点中是否只有一个节点的入度为零，终止节点中的入度是否都为一即可。

代码实现：

#include <cstdlib>
#include <cstdio>
#include <iostream>

using namespace std;
#define N 10010
int tree[N];//用于存储各节点的入度

typedef struct
{
    int from;//起始节点
    int to;//终止节点

}Edge;

int count=0;
Edge edge[N];//用于存储树中的有向边

int main()
{
    int n,m;
    int i;
    int k;
    while(~scanf("%d%d",&n,&m)&&n!=-1)
    {
        k=0;
        count++;
        if(m==n&&n==0)
        {
            cout<<"Case "<<count<<" is a tree."<<endl;//不存在节点也为一课树
            continue;
        }
        edge[k].from=n;
        edge[k].to=m;
        k++;
        for(i=0;i<N;i++)
            tree[i]=0;
        ++tree[m];
        while(~scanf("%d%d",&n,&m)&&n!=0&&m!=0)
        {
            ++tree[m];
            edge[k].from=n;
            edge[k].to=m;
            k++;
        }
        bool flog=false;
        for(i=0;i<k;i++)
        {
            if((tree[edge[i].from]==0&&flog==false))
            {
                flog=true;
            }
            else if(tree[edge[i].from]==0&&flog==true)
            {
                break;
            }//上述条件语句用于判断是否只有一个起始节点的入度为零
            if(tree[edge[i].to]>1)
            {
                break;
            }//用于判断终止节点的度是否都为一
        }
        if(flog&&i>=k)
        {
            cout<<"Case "<<count<<" is a tree."<<endl;
        }
        else
            cout<<"Case "<<count<<" is not a tree."<<endl;
    }
    return 0;
}

结果展示: