zoukankan      html  css  js  c++  java
  • N

    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.
    #include <iostream>
    #include <string>
    #include<cstring>
    #include<cstdio>
    #include <vector>
    #include <algorithm>
    using namespace std;
    const int MAXN = 50005;
    int pre[MAXN],tmp[MAXN],cnt=0;
    int find(int x)
    {
        if(pre[x]==-1)
            return x;
        else
            return pre[x] = find(pre[x]);
    };
    int main()
    {
        int x,y;
        bool f = false;
        memset(pre,-1,sizeof(pre));
        int cas=1;
        while(scanf("%d%d",&x,&y))
        {
            if(x==-1&&y==-1) break;
            if(x==0&&y==0)
            {
                int num = 0,ft = find(tmp[0]);
                if(!f)
                {
                    for(int i=0;i<cnt;i++)
                    {
                        if(find(tmp[i])!=ft)
                            num++;
                    }
                    if(num) f = true;
                }
                if(!f)
                    printf("Case %d is a tree.
    ",cas);
                else
                    printf("Case %d is not a tree.
    ",cas);
                memset(pre,-1,sizeof(pre));
                cas++;
                f = false;
                cnt = 0;
                continue;
            }
            int fx=find(x),fy=find(y);
            tmp[cnt++] = x;
            tmp[cnt++] = y;
            if(fx!=fy)
                pre[fy] = fx;
            else
                f = true;
        }
        return 0;
    }
  • 相关阅读:
    Leetcode 18. 4Sum
    Leetcode 15. 3Sum
    Leetcode 16. 3Sum Closest
    String类型的理解
    求字符串中某个字符出现的次数
    用StringBuilder来实现经典的反转问题
    String/StringBuilder 类 用对象数组实现登录注册功能
    String/StringBuilder 类 统计字符串中字符出现的次数
    String/StringBuilder 类 判断QQ号码
    C++ 面向对象: I/O对象的应用
  • 原文地址:https://www.cnblogs.com/joeylee97/p/6615921.html
Copyright © 2011-2022 走看看