Is It A Tree?【 POJ - 1308】

最新推荐文章于 2022-02-15 15:15:00 发布

原创最新推荐文章于 2022-02-15 15:15:00 发布 · 173 阅读

CC 4.0 BY-SA版权

本文介绍了一种通过并查集或直接利用树的定义来判断一组节点连接是否构成树的算法。该算法适用于输入一系列节点及它们之间的连接关系，并确定这些连接是否符合树的特性。

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.

题意：

判断这是不是一棵树，可以用并查集，也可以根据树的定义做。这道题一开始习惯性把清零函数写在最上边，造成了察觉不到的错误。

代码：

#include<cstdio>
#include<cstring>

int w[100100],n,m,k,sum;
bool s[100100];

int main()
{
    int kk=1;
    while(scanf("%d%d",&n,&m)&&(n!=-1||m!=-1))
    {
        if(n==0&&m==0)
        {
            printf("Case %d is a tree.\n",kk);
            kk++;
            continue;
        }
        k=0;
        sum=0;
        bool flag=true;
        memset(w,0,sizeof(w));
        memset(s,false,sizeof(s));
        while(!(n==0&&m==0))
        {
            if(flag==true)
            {
                if(w[m]>0)
                    flag=false;
                if(s[n]==false)
                {
                    k++;
                    s[n]=true;
                }
                if(s[m]==false)
                {
                    k++;
                    s[m]=true;
                }
                sum++;
                w[m]=n;
            }
            scanf("%d%d",&n,&m);
        }
        if(k==sum+1 && flag==true)
            printf("Case %d is a tree.\n",kk);
        else
            printf("Case %d is not a tree.\n",kk);
        kk++;
    }
    return 0;
}