Aantal parallellogrammen in een vlak
Gegeven enkele punten op een vlak die verschillend zijn en geen drie ervan op dezelfde lijn liggen. We moeten het aantal parallellogrammen vinden met de hoekpunten als de gegeven punten. Voorbeelden:
Input : points[] = {(0 0) (0 2) (2 2) (4 2) (1 4) (3 4)} Output : 2 Two Parallelograms are possible by choosing above given point as vertices which are shown in below diagram. We kunnen dit probleem oplossen door de speciale eigenschap van parallellogrammen te gebruiken, namelijk dat diagonalen van een parallellogram elkaar in het midden snijden. Dus als we zo'n middelpunt krijgen dat het middelpunt is van meer dan één lijnsegment, dan kunnen we concluderen dat een parallellogram nauwkeuriger bestaat als een middelpunt x keer voorkomt, dan kunnen diagonalen van mogelijke parallellogrammen worden gekozen X C 2 manieren, d.w.z. er zullen x*(x-1)/2 parallellogrammen zijn die overeenkomen met dit specifieke middelpunt met een frequentie x. We herhalen dus alle paar punten en berekenen hun middelpunt en verhogen de frequentie van het middelpunt met 1. Aan het einde tellen we het aantal parallellogrammen volgens de frequentie van elk afzonderlijk middelpunt, zoals hierboven uitgelegd. Omdat we alleen de frequentie van de middelpuntdeling door 2 nodig hebben, wordt deze voor de eenvoud genegeerd bij het berekenen van het middelpunt.
CPP // C++ program to get number of Parallelograms we // can make by given points of the plane #include using namespace std ; // Returns count of Parallelograms possible // from given points int countOfParallelograms ( int x [] int y [] int N ) { // Map to store frequency of mid points map < pair < int int > int > cnt ; for ( int i = 0 ; i < N ; i ++ ) { for ( int j = i + 1 ; j < N ; j ++ ) { // division by 2 is ignored to get // rid of doubles int midX = x [ i ] + x [ j ]; int midY = y [ i ] + y [ j ]; // increase the frequency of mid point cnt [ make_pair ( midX midY )] ++ ; } } // Iterating through all mid points int res = 0 ; for ( auto it = cnt . begin (); it != cnt . end (); it ++ ) { int freq = it -> second ; // Increase the count of Parallelograms by // applying function on frequency of mid point res += freq * ( freq - 1 ) / 2 ; } return res ; } // Driver code to test above methods int main () { int x [] = { 0 0 2 4 1 3 }; int y [] = { 0 2 2 2 4 4 }; int N = sizeof ( x ) / sizeof ( int ); cout < < countOfParallelograms ( x y N ) < < endl ; return 0 ; }
Java /*package whatever //do not write package name here */ import java.io.* ; import java.util.* ; public class GFG { // Returns count of Parallelograms possible // from given points public static int countOfParallelograms ( int [] x int [] y int N ) { // Map to store frequency of mid points HashMap < String Integer > cnt = new HashMap <> (); for ( int i = 0 ; i < N ; i ++ ) { for ( int j = i + 1 ; j < N ; j ++ ) { // division by 2 is ignored to get // rid of doubles int midX = x [ i ] + x [ j ] ; int midY = y [ i ] + y [ j ] ; // increase the frequency of mid point String temp = String . join ( ' ' String . valueOf ( midX ) String . valueOf ( midY )); if ( cnt . containsKey ( temp )){ cnt . put ( temp cnt . get ( temp ) + 1 ); } else { cnt . put ( temp 1 ); } } } // Iterating through all mid points int res = 0 ; for ( Map . Entry < String Integer > it : cnt . entrySet ()) { int freq = it . getValue (); // Increase the count of Parallelograms by // applying function on frequency of mid point res = res + freq * ( freq - 1 ) / 2 ; } return res ; } public static void main ( String [] args ) { int [] x = { 0 0 2 4 1 3 }; int [] y = { 0 2 2 2 4 4 }; int N = x . length ; System . out . println ( countOfParallelograms ( x y N )); } } // The code is contributed by Nidhi goel.
Python3 # python program to get number of Parallelograms we # can make by given points of the plane # Returns count of Parallelograms possible # from given points def countOfParallelograms ( x y N ): # Map to store frequency of mid points cnt = {} for i in range ( N ): for j in range ( i + 1 N ): # division by 2 is ignored to get # rid of doubles midX = x [ i ] + x [ j ]; midY = y [ i ] + y [ j ]; # increase the frequency of mid point if (( midX midY ) in cnt ): cnt [( midX midY )] += 1 else : cnt [( midX midY )] = 1 # Iterating through all mid points res = 0 for key in cnt : freq = cnt [ key ] # Increase the count of Parallelograms by # applying function on frequency of mid point res += freq * ( freq - 1 ) / 2 return res # Driver code to test above methods x = [ 0 0 2 4 1 3 ] y = [ 0 2 2 2 4 4 ] N = len ( x ); print ( int ( countOfParallelograms ( x y N ))) # The code is contributed by Gautam goel.
C# using System ; using System.Collections.Generic ; public class GFG { // Returns count of Parallelograms possible // from given points public static int CountOfParallelograms ( int [] x int [] y int N ) { // Map to store frequency of mid points Dictionary < string int > cnt = new Dictionary < string int > (); for ( int i = 0 ; i < N ; i ++ ) { for ( int j = i + 1 ; j < N ; j ++ ) { // division by 2 is ignored to get // rid of doubles int midX = x [ i ] + x [ j ]; int midY = y [ i ] + y [ j ]; // increase the frequency of mid point string temp = string . Join ( ' ' midX . ToString () midY . ToString ()); if ( cnt . ContainsKey ( temp )) { cnt [ temp ] ++ ; } else { cnt . Add ( temp 1 ); } } } // Iterating through all mid points int res = 0 ; foreach ( KeyValuePair < string int > it in cnt ) { int freq = it . Value ; // Increase the count of Parallelograms by // applying function on frequency of mid point res += freq * ( freq - 1 ) / 2 ; } return res ; } public static void Main ( string [] args ) { int [] x = { 0 0 2 4 1 3 }; int [] y = { 0 2 2 2 4 4 }; int N = x . Length ; Console . WriteLine ( CountOfParallelograms ( x y N )); } }
JavaScript // JavaScript program to get number of Parallelograms we // can make by given points of the plane // Returns count of Parallelograms possible // from given points function countOfParallelograms ( x y N ) { // Map to store frequency of mid points // map int> cnt; let cnt = new Map (); for ( let i = 0 ; i < N ; i ++ ) { for ( let j = i + 1 ; j < N ; j ++ ) { // division by 2 is ignored to get // rid of doubles let midX = x [ i ] + x [ j ]; let midY = y [ i ] + y [ j ]; // increase the frequency of mid point let make_pair = [ midX midY ]; if ( cnt . has ( make_pair . join ( '' ))){ cnt . set ( make_pair . join ( '' ) cnt . get ( make_pair . join ( '' )) + 1 ); } else { cnt . set ( make_pair . join ( '' ) 1 ); } } } // Iterating through all mid points let res = 0 ; for ( const [ key value ] of cnt ) { let freq = value ; // Increase the count of Parallelograms by // applying function on frequency of mid point res = res + Math . floor ( freq * ( freq - 1 ) / 2 ); } return res ; } // Driver code to test above methods let x = [ 0 0 2 4 1 3 ]; let y = [ 0 2 2 2 4 4 ]; let N = x . length ; console . log ( countOfParallelograms ( x y N )); // The code is contributed by Gautam goel (gautamgoel962)
Uitvoer
2
Tijdcomplexiteit: Op 2 logn) terwijl we door twee lussen doorlopen tot n en ook een kaart gebruiken die logn nodig heeft.
Hulpruimte: Op)
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