05-树8 File Transfer (25 分)

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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 (2N10?4??), 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.


特意查了查c++的 二维数组 怎么声明 ,函数 参数是二维数组的情况怎么做,返回是一维数组怎么弄,
查到用的是指针.就那么用吧,估计就该这样吧.

c#的代码不能满分
c++的可以

using System;
using System.Collections.Generic;
using System.Threading;
using System.Threading.Tasks;
using System.Diagnostics;
using System.Net;
using System.Text;
using System.Xml;

class T
{



    public class MyItem
    {
        public int Value;
        public int PIndex;

    }
    static void Main(string[] args)
    {

        List<MyItem> list = new List<MyItem>();

        var count = int.Parse(Console.ReadLine());
        for (int i = 1; i < count + 1; i++)
        {
            list.Add(new MyItem() { Value = i, PIndex = -1 });
        }
        AAA:

        var a = Console.ReadLine();
        if (a != "S")
        {
            var item = a.Split( );
            var v1 = 查找(list, item[1]);
            var v2 = 查找(list, item[2]);
            switch (item[0])
            {
                case "C":

                    if (v1.Value == v2.Value)
                    {
                        Console.WriteLine("yes");
                    }
                    else
                    {
                        Console.WriteLine("no");
                    }
                    break;
                case "I":


                    if (v1.PIndex <= v2.PIndex)
                    {

                        v1.PIndex += v2.PIndex;
                        v2.PIndex = v1.Value;

                    }
                    else
                    {
                        v2.PIndex += v1.PIndex;
                        v1.PIndex = v2.Value;
                    }


                    break;

                default:
                    break;
            }
            goto AAA;
        }
        else
        {
            var v = list.FindAll(ax => ax.PIndex < 0);
            if (v.Count == 1)
            {
                Console.WriteLine("The network is connected.");
            }
            else
            {
                Console.WriteLine($"There are {v.Count} components.");
            }
            return;
        }




    }

    private static MyItem 查找(List<MyItem> list, string item)
    {
        var myItem = list[int.Parse(item)-1];


        if (myItem != null)
        {
            while (myItem.PIndex > 0)
            {
                myItem = list.Find(x => x.Value == myItem.PIndex);
            }
            return myItem;
        }
        else
        {
            return null;
        }
    }
}

 

 
// C1.cpp : 此文件包含 "main" 函数。程序执行将在此处开始并结束。
//

//#include "pch.h"
#include <iostream>
using namespace std;



int* find(int **list, int i1);

int ccc = 0;
int main()
{

    int **list;


    std::cin >> ccc;

    list = new int*[ccc];

    for (int i = 0; i < ccc; i++)
    {
        list[i] = new int[2];
        list[i][0] = i + 1;
        list[i][1] = -1;
    }

AAA:

    char a = a;
    std::cin >> a;

    if (a != S)
    {
        int i1;
        int i2;
        std::cin >> i1;
        std::cin >> i2;


        int *v1 = find(list, i1);
        int *v2 = find(list, i2);
        switch (a)
        {
        case C:

            if (v1[0] == v2[0])
            {
                cout << ("yes
");
            }
            else
            {
                cout << ("no
");
            }
            break;
        case I:


            if (v1[1] <= v2[1])
            {

                v1[1] += v2[1];
                v2[1] = v1[0];

            }
            else
            {
                v2[1] += v1[1];
                v1[1] = v2[0];
            }


            break;

        default:
            break;
        }
        goto AAA;
    }
    else
    {
        int ci = 0;
        for (int i = 0; i < ccc; i++)
        {
            if (list[i][1] < 0)
            {
                ci++;
            }

        }
        if (ci == 1)
        {
            cout << ("The network is connected.
");
        }
        else
        {

            cout << "There are " << ci << " components.
";
        }

    }

    return 0;

}
int * find(int **list, int i1)
{
    int i = i1 - 1;
    for (; i < ccc; i++)
    {
        if (list[i][0] == i1)
        {
            break;
        }

    }
    while (list[i][1] > 0)
    {
        i = list[i][1] - 1;
    }

    return  list[i];
}

 

 









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