zoukankan      html  css  js  c++  java
  • hdu 1325 Is It A Tree? (并查集)

    Is It A Tree?

    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.

    这里写图片描述 这里写图片描述 这里写图片描述

    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 lineCase 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<stdio.h>
    #include<string.h>
    int s[100010];
    int z[100010];
    //int v[100010];
    int flag;
    int find(int x)
    {
        if(s[x]==x)
        return x;
        else
        return s[x]=find(s[x]);
    }
    
    
    void mix(int a,int b )
    {
        int x,y;
        x=find(a );
        y=find(b );
        if(x==y||y!=b)
            flag=0;
        else
            s[y]=x;
    }
    
    int main()
    {
        int a,b;
        int Case=1;
        while(scanf("%d%d",&a,&b)!=EOF)
        {
            if(a<0&&b<0)
            break;
            else
            {
    
                for(int i=1;i<=100001;i++)
                s[i]=i;
                flag=1;
                memset(z,0,sizeof(z));
        //      memset
                int MAX=0;
                if(a==0&&b==0)
                printf("Case %d is a tree.
    ",Case++);
                else
                {
    
                    z[a]=1;
                    z[b]=1;
                    mix(a,b);
                    if(a>MAX)
                    MAX=a;
                    if(b>MAX)
                    MAX=b;
                    int k=1; 
                    while(scanf("%d%d",&a,&b)!=EOF)
                    {
                        if(a==0&&b==0)
                        break;
                        else
                        {
                            z[a]=1;
                            z[b]=1;
                            mix(a,b);
                            if(a>MAX)
                            MAX=a;
                            if(b>MAX)
                            MAX=b;
                            k++;
                        }
                    }
                    int sum=0,ans=0; 
                    for(int i=1;i<=MAX;i++)
                    {
                        sum+=z[i];
                        if(s[i]==i&&z[i]==1)
                        ans++;
                    }
    
                    if(k+1==sum&&flag&&ans==1)
                    printf("Case %d is a tree.
    ",Case++);
                    else
                    printf("Case %d is not a tree.
    ",Case++);
    
                }
            }
        }
        return 0;
    }
  • 相关阅读:
    后端开发应该掌握的 Redis 基础
    Code Review有什么好处?
    对不起,你那不叫努力,叫重复劳动
    老鸟程序员才知道的40个小技巧
    单例模式基础笔记
    最受IT公司欢迎的50款开源软件
    硬件:关于路由器、交换机、宽带猫的几个问题
    硬件:宽带猫(光猫)的基础知识
    python selenium模块使用出错解决,Message: ‘geckodriver’ executable needs to be in PATH
    python+selenium如何定位页面的元素,的几种定位元素的方法。
  • 原文地址:https://www.cnblogs.com/-xiangyang/p/9220259.html
Copyright © 2011-2022 走看看