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mojhsa [17]
3 years ago
7

Assume that

Computers and Technology
1 answer:
gulaghasi [49]3 years ago
6 0

Answer:

Explanation:

Create a min-heap of size v (the number of vertices), each of which contains the vertex number and the distance from the root vertex (which is the source vertex).

So, the distance on the root vertex would be 0. And let the distances on all other nodes be infinite(since they will be updated later).

Until the min-heap gets empty, do the following

(i) Extract that vertex from the min-heap, which has the minimum distance value, using Extract-Min operation, which takes O(logV). Lets name it u

(ii) Now, for every adjacent vertex of u, say v, check if v is in min-heap or not. If yes, and if the distance value of u plus the edge weight u-v is less than the distance value of v, then update the distance value of v.

Running time of O(ElogV) is obtained, as the Extract-Min operation which takes O(logV), is performed at most E times, i.e. the number of edges times.

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tatyana61 [14]

Answer:

i can't see this

Explanation:

5 0
3 years ago
How many different integers can be represented with a 4-digit number in base 13?
AnnyKZ [126]
That would be 13^4, or 13*13*13*13 = 28561
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Which element can be changed using the Print pane? Check all that apply.
Ronch [10]

Answer:

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D

E

Explanation:

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Read 2 more answers
1. Print out the string length of s1 2. Loop through characters in s2 with charAt() and display reversed string 3. Compare s2, s
konstantin123 [22]

Answer:

See explaination

Explanation:

public class StringLab9 {

public static void main(String args[]) {

char charArray[] = { 'C', 'O', 'S', 'C', ' ', '3', '3', '1', '7', ' ', 'O', 'O', ' ', 'C', 'l', 'a', 's', 's' };

String s1 = new String("Objected oriented programming language!");

String s2 = "COSC 3317 OO class class";

String s3 = new String(charArray);

// To do 1: print out the string length of s1

System.out.println(s1.length());

// To do 2: loop through characters in s2 with charAt and display reversed

// string

for (int i = s2.length() - 1; i >= 0; --i)

System.out.print(s2.charAt(i));

System.out.println();

// To do 3: compare s2, s3 with compareTo(), print out which string (s2 or s3)

// is

// greater than which string (s2 or s3), or equal, print the result out

if (s2.compareTo(s3) == 0)

System.out.println("They are equal");

else if (s2.compareTo(s3) > 0)

System.out.println("s2 is greater");

else

System.out.println("s3 is greater");

// To do 4: Use the regionMatches to compare s2 and s3 with case sensitivity of

// the first 8 characters.

// and print out the result (match or not) .

if (s2.substring(0, 8).compareTo(s3.substring(0, 8)) == 0)

System.out.println("They matched");

else

System.out.println("They DONT match");

// To do 5: Find the location of the first character 'g' in s1, print it out

int i;

for (i = 0; i < s2.length(); ++i)

if(s2.charAt(i)=='g')

break;

System.out.println("'g' is present at index " + i);

// To do 6: Find the last location of the substring "class" from s2, print it

// out

int index = 0, ans = 0;

String test = s2;

while (index != -1) {

ans = ans + index;

index = test.indexOf("class");

test = test.substring(index + 1, test.length());

}

System.out.println("Last location of class in s2 is: " + (ans + 1));

// To do 7: Extract a substring from index 4 up to, but not including 8 from

// s3, print it out

System.out.println(s3.substring(4, 8));

} // end main

} // end class StringLab9

7 0
3 years ago
Choose the correct item. Check your answers in the text.<br>3​
RSB [31]

Answer:

You should do that

Explanation:

You didnt say anything

5 0
3 years ago
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