Lygiagretainių skaičius plokštumoje
Duoti kai kurie plokštumos taškai, kurie yra skirtingi ir nė vienas iš jų nėra vienoje tiesėje. Turime rasti lygiagrečių skaičių, kurių viršūnės yra nurodyti taškai. Pavyzdžiai:
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. Šį uždavinį galime išspręsti pasinaudoję specialia lygiagretainio savybe, kad lygiagretainio įstrižainės kerta viena kitą per vidurį. Taigi, jei gausime tokį vidurio tašką, kuris yra daugiau nei vienos tiesės atkarpos vidurio taškas, galime daryti išvadą, kad lygiagretainis egzistuoja tiksliau, jei vidurio taškas atsiranda x kartų, tada galimų lygiagretainių įstrižaines galima pasirinkti x C 2 būdai, ty bus x*(x-1)/2 lygiagretainiai, atitinkantys šį konkretų vidurinį tašką, kurio dažnis x. Taigi kartojame visas taškų poras ir apskaičiuojame jų vidurinį tašką ir padidiname vidurinio taško dažnį 1. Pabaigoje suskaičiuojame lygiagretainių skaičių pagal kiekvieno atskiro vidurinio taško dažnį, kaip paaiškinta aukščiau. Kadangi mums tiesiog reikia vidurinio taško padalijimo iš 2 dažnio, skaičiuojant vidurinį tašką paprastumo dėlei nepaisoma.
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)
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Laiko sudėtingumas: O (n 2 logn), nes kartojame per dvi kilpas iki n ir taip pat naudojame žemėlapį, kuriam reikia logn.
Pagalbinė erdvė: O(n)
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