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blondinia [14]
3 years ago
14

Several families are planning a shared car trip on scenic drives in New Hampshire's White Mountains. To minimize the possibility

of any quarrels, they want to assign individuals to cars so that no two members of a family are in the same car. Explain how to formulate this problem as a network flow problem.
Computers and Technology
1 answer:
Rus_ich [418]3 years ago
5 0

Answer:

Following are the response to the given question:  

Explanation:

Build a spring, sink, vertices, and vertices for each car for a household. Every unit in the stream is a human. Attach the source from each vertical of a family with such a capacity line equivalent to the family size; this sets the number of members in each household. Attach every car vertices to the sink with the edge of the car's passenger belt; this assures the correct number of people for every vehicle. Connecting every vertex in your household to any vertex in your vehicle with a capacity 1 border guarantees that one family member joins a single car. The link between both the acceptable allocation of people to vehicles as well as the maximum flow inside the graph seems clear to notice.

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Answer:

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Explanation:

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How do you recognize substances that have gone through physical or chemical changes?
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Make sure to read all of these specifications carefully. Write a function, named array_shifter, that accepts an array of doubles
rosijanka [135]

Answer:

import java.util.Arrays;

public class num2 {

   public static void main(String[] args) {

//Create and initialize the first array        

double [ ]arr = {3.5, 5.6, 4.5, 6.7};

// Get the length of the array

       int n = arr.length;

      //Call the method Array Shifter inside the output statement

       System.out.println(Arrays.toString(array_shifter(arr,n)));

   }

   static double [] array_shifter(double [ ] doubleArray, int n){

       double [] narr = new double[n+1];

       //Loop through from index 1

       for(int i =0; i<narr.length-1; i++){

           narr[i] = doubleArray[i];

       }

       //Put a new element at the last index

       narr[narr.length-1] = 1.1;

       return narr;

   }

}

Explanation:

This is implemented in Java

Read detailed comments in the solution

The method array_shifter creates a new array that is one element larger than the array it received as argument and returns it

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Design and implement a class dayType that implements the day of the week in a program. The class dayType should store the day, s
Afina-wow [57]

The code is implemented based on the given operations.

Explanation:

#include <iostream>

#include <string>

using namespace std;

class dayType

{ private:

 string day[7];

 string presentDay;

 int numofDays;

public:

 void setDay(string freshDay);

 void printDay() const;

 int showDay(int &day);

 int nextDay(int day);

 int prevDay(int day) const;

 int calcDay(int day, int numofDays);    

 dayType()

 {

  day[0] = "Sunday";

  day[1] = "Monday";

  day[2] = "Tuesday";

  day[3] = "Wednesday";

  day[4] = "Thursday";

  day[5] = "Friday";

  day[6] = "Saturday";

  presentDay = day[0];

  numofDays = 0;

 };

 ~dayType();

};

#endif

#include "dayType.h"

void dayType::setDay(string freshDay)

{

  presentDay = freshDay;

}

void dayType::printDay()

{

  cout << "Day chosen is " << presentDay << endl;

}

int dayType::showDay(int& day)

{

  return day;

}

int dayType::nextDay(int day)

{

day = day++;

if (day > 6)

 day = day % 7;

switch (day)

{

case 0: cout << "The successive day is Sunday";

 break;

case 1: cout << "The successive day is Monday";

 break;

case 2: cout << "The successive day is Tuesday";

 break;

case 3: cout << "The successive day is Wednesday";

 break;

case 4: cout << "The successive day is Thursday";

 break;

case 5: cout << "The successive day is Friday";

 break;

case 6: cout << "The successive day is Saturday";

 break;

}

cout << endl;

return day;

}

 

int dayType::prevDay(int day)

{

day = day--;

switch (day)

{

case -1: cout << "The before day is Saturday.";

 break;

case 0: cout << "The before day is Saturday.";

 break;

case 1: cout << "The before day is Saturday.";

 break;

case 2: cout << "The before day is Saturday.";

 break;

case 3: cout << "The before day is Saturday.";

 break;

case 4: cout << "The before day is Saturday.";

 break;

case 5: cout << "The before day is Saturday.";

 break;

default: cout << "The before day is Saturday.";

}

cout << endl;

return day;

}

int dayType::calcDay(int addDays, int numofDays)

{

addDay = addDays + numofDays;

if (addDay > 6)

 addDay = addDay % 7;

switch(addDay)

{

case 0: cout << "The processed day is Sunday.";

 break;

case 1: cout << "The processedday is Monday.";

 break;

case 2: cout << "The processedday is Tuesday.";

 break;

case 3: cout << "The processedday is Wednesday.";

 break;

case 4: cout << "The processedday is Thursday.";

 break;

case 5: cout << "The processedday is Friday.";

 break;

case 6: cout << "The processedday is Saturday.";

 break;

default: cout << "Not valid choice.";

}

cout << endl;

return addDays;

}

4 0
3 years ago
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