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GalinKa [24]
3 years ago
7

Suppose that, in addition to edge capacities, a flow network has vertex capacities. That is each vertex vv has a limit l(v)l(v)

on how much flow can pass though vv. Show how to transform a flow network G = (V, E)G=(V,E) with vertex capacities into an equivalent flow network G' = (V', E')G ′ =(V ′ ,E ′ ) without vertex capacities, such that a maximum flow in G'G ′ has the same value as a maximum flow in GG. How many vertices and edges does G'G ′ have?
Computers and Technology
1 answer:
andrey2020 [161]3 years ago
6 0

Answer:

there would be 14 vertices and 7 edges

Explanation:

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Write a while loop that prints
Nana76 [90]

Answer:

The program to this question as follows:

Program:

#include <iostream> //defining header file

using namespace std;

int main() //defining main method

{

int squ=0,n=0; //defining variable

cout<<"Square between 0 to 100 :"; //message

while(n<100) //loop for calculate Square

{

n=squ*squ; //holing value in n variable

cout<<n<<" "; //print Square

squ++; //increment value by 1

}

cout<<endl; //for new line

n=1; //change the value of n

cout<<"Positive number, which is divisible by 10: "; //message

while (n< 100) //loop for check condition

{

if(n%10==0) //check value is divisible by 10

{

cout<<n<<" ";//print value

}

n++; //increment value of n by 1

}

cout<<endl; //for new line

cout<<"Powers of two less than n: "; //message

n=1; //holing value in n

while (n< 100) //loop for check condition

{

cout<<n<<" ";//print value

n=n*2; //calculate value

}

return 0;

}

Output:

Square between 0 to 100 :0 1 4 9 16 25 36 49 64 81 100  

Positive number, which is divisible by 10: 10 20 30 40 50 60 70 80 90  

Powers of two less than n: 1 2 4 8 16 32 64  

Explanation:

In the above program, two integer variable "squ and n" is defined, then three while loop is declared, which is used to calculate different values, that can be described as follows:

  • In the first while loop both "n and squ" variable is used, in which n is used for check range and "squ" is used to calculate the square between 1 to 100.
  • The second, while it is used to calculate the positive number, which is divisible by 10, in this only n variable is used, that calculates the value and check its range.
  • Then the last while loop is used which is used to calculate the double of the number, which is between 1 to 100 range.  
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Answer:

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