Contagem de paralelogramos em um plano
Dados alguns pontos em um plano que são distintos e não há três deles na mesma linha. Precisamos encontrar o número de paralelogramos com os vértices como pontos dados. Exemplos:
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. Podemos resolver este problema utilizando uma propriedade especial dos paralelogramos: as diagonais de um paralelogramo intersectam-se no meio. Portanto, se obtivermos um ponto médio que é o ponto médio de mais de um segmento de linha, podemos concluir que um paralelogramo existe com mais precisão se um ponto médio ocorrer x vezes, então as diagonais de paralelogramos possíveis podem ser escolhidas em x C 2 maneiras, ou seja, haverá x*(x-1)/2 paralelogramos correspondentes a este ponto médio específico com uma frequência x. Então, iteramos sobre todos os pares de pontos e calculamos seu ponto médio e aumentamos a frequência do ponto médio em 1. No final, contamos o número de paralelogramos de acordo com a frequência de cada ponto médio distinto, conforme explicado acima. Como precisamos apenas da frequência do ponto médio, a divisão por 2 é ignorada ao calcular o ponto médio para simplificar.
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)
Saída
2
Complexidade de tempo: Sobre 2 logn), pois estamos iterando através de dois loops até n e usando também um mapa que leva logn.
Espaço Auxiliar: Sobre)
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