• 05-树8 File Transfer (25 分)


    We have a network of computers and a list of bi-directional connections. Each of these connections allows a file transfer from one computer to another. Is it possible to send a file from any computer on the network to any other?

    Input Specification:

    Each input file contains one test case. For each test case, the first line contains N (2), the total number of computers in a network. Each computer in the network is then represented by a positive integer between 1 and N. Then in the following lines, the input is given in the format:

    I c1 c2  
    

    where I stands for inputting a connection between c1 and c2; or

    C c1 c2    
    

    where C stands for checking if it is possible to transfer files between c1 and c2; or

    S
    

    where S stands for stopping this case.

    Output Specification:

    For each C case, print in one line the word "yes" or "no" if it is possible or impossible to transfer files between c1 and c2, respectively. At the end of each case, print in one line "The network is connected." if there is a path between any pair of computers; or "There are k components." where k is the number of connected components in this network.

    Sample Input 1:

    5
    C 3 2
    I 3 2
    C 1 5
    I 4 5
    I 2 4
    C 3 5
    S
    

    Sample Output 1:

    no
    no
    yes
    There are 2 components.
    

    Sample Input 2:

    5
    C 3 2
    I 3 2
    C 1 5
    I 4 5
    I 2 4
    C 3 5
    I 1 3
    C 1 5
    S
    

    Sample Output 2:

    no
    no
    yes
    yes
    The network is connected.
    #include<cstdio>
    const int maxn = 1010;
    
    void Initialization( int n);
    void Input_connection();
    int FindFather(int x);
    void Check_connection();
    void Check_network(int n);
    int arr[maxn];
    
    int main()
    {
        int n;
        char in;
        
        scanf("%d",&n);
        
        Initialization(n);
        
        do
        {
            getchar();
            scanf("%c",&in);
            switch(in)
            {
                
                case 'I':
                {
                    Input_connection();
                    break;
                }
                case 'C':
                {
                    Check_connection();
                    break;
                }
                case 'S':
                {
                    Check_network(n);
                    break;
                }
            }
        }while (in != 'S');
        return 0;
    
    }
    
    void Initialization( int n)
    {
        for (int i = 0; i <= n; i++)
        {
            arr[i] = i;    
        }    
    }
    
    void  Input_connection()
    {
        int u,v;
        scanf("%d %d",&u,&v);
        int faA = FindFather(u);
        int faB = FindFather(v);
        if (faA != faB)
        {
            arr[faA] = faB;
        }
    }
    
    int FindFather(int x)
    {
        if (arr[x] == x)
        {
            return x;
        }
        else
        {
            return arr[x] = FindFather(arr[x]);
        }
    }
    
    void Check_connection()
    {
        int u,v;
        scanf("%d %d",&u,&v);
        int faA = FindFather(u);
        int faB = FindFather(v);
        if (faA != faB)
        {
            printf("no
    ");
        }    
        else
        {
            printf("yes
    ");
        }
    }
    
    void Check_network(int n)
    {
        int cnt = 0;
        for (int i = 1; i <= n; i++)
        {
            if (arr[i] == i)
            {
                cnt++;
            }
        }
        
        if (1 == cnt)
        {
            printf("The network is connected.");
        }
        else
        {
            printf("There are %d components.",cnt);
        }
    }
  • 相关阅读:
    Vue 监听子组件事件
    延时队列
    AES加密
    centos7.9 iftop 工具源码安装
    angular pass get paragrams by router
    Android chrome console in PC
    powershell 运行带路径的exe
    win下 nrm ls报错
    windows10 安装 node 16 解决node-sass node-gyp报错
    位图和布隆过滤器
  • 原文地址:https://www.cnblogs.com/wanghao-boke/p/11761751.html
Copyright © 2020-2023  润新知