Tulosta parien enimmäispituus
Sinulle annetaan n paria numeroita. Jokaisessa parissa ensimmäinen numero on aina pienempi kuin toinen numero. Pari (c d) voi seurata toista paria (a b), jos b < c. Chain of pairs can be formed in this fashion. Find the longest chain which can be formed from a given set of pairs. Esimerkkejä:
Input: (5 24) (39 60) (15 28) (27 40) (50 90) Output: (5 24) (27 40) (50 90) Input: (11 20) {10 40) (45 60) (39 40) Output: (11 20) (39 40) (45 60) sisään edellinen viesti, jonka olemme keskustelleet parien enimmäispituusketjun ongelmasta. Viesti koski kuitenkin vain enimmäiskokoketjun pituuden löytämiseen liittyvää koodia, mutta ei enimmäiskokoketjun rakentamista. Tässä viestissä keskustelemme kuinka rakentaa itse parien enimmäispituusketju. Ajatuksena on ensin lajitella annetut parit niiden ensimmäisen elementin kasvavaan järjestykseen. Olkoon arr[0..n-1] parien syötetaulukko lajittelun jälkeen. Määrittelemme vektorin L siten, että L[i] on itse vektori, joka tallentaa arr[0..i]-parien maksimipituusketjun, joka päättyy arr[i]:iin. Siksi indeksille i L[i] voidaan kirjoittaa rekursiivisesti muodossa -
L[0] = {arr[0]} L[i] = {Max(L[j])} + arr[i] where j < i and arr[j].b < arr[i].a = arr[i] if there is no such j Esimerkiksi (5 24) (39 60) (15 28) (27 40) (50 90)
L[0]: (5 24) L[1]: (5 24) (39 60) L[2]: (15 28) L[3]: (5 24) (27 40) L[4]: (5 24) (27 40) (50 90)
Huomaa, että parien lajittelu on tehty, koska meidän on löydettävä parin maksimipituus ja tilauksella ei ole tässä väliä. Jos emme lajittele, saamme parit kasvavassa järjestyksessä, mutta ne eivät ole mahdollisimman suuria pareja. Alla ylläolevan idean toteutus -
C++ /* Dynamic Programming solution to construct Maximum Length Chain of Pairs */ #include using namespace std ; struct Pair { int a ; int b ; }; // comparator function for sort function int compare ( Pair x Pair y ) { return x . a < y . a ; } // Function to construct Maximum Length Chain // of Pairs void maxChainLength ( vector < Pair > arr ) { // Sort by start time sort ( arr . begin () arr . end () compare ); // L[i] stores maximum length of chain of // arr[0..i] that ends with arr[i]. vector < vector < Pair > > L ( arr . size ()); // L[0] is equal to arr[0] L [ 0 ]. push_back ( arr [ 0 ]); // start from index 1 for ( int i = 1 ; i < arr . size (); i ++ ) { // for every j less than i for ( int j = 0 ; j < i ; j ++ ) { // L[i] = {Max(L[j])} + arr[i] // where j < i and arr[j].b < arr[i].a if (( arr [ j ]. b < arr [ i ]. a ) && ( L [ j ]. size () > L [ i ]. size ())) L [ i ] = L [ j ]; } L [ i ]. push_back ( arr [ i ]); } // print max length vector vector < Pair > maxChain ; for ( vector < Pair > x : L ) if ( x . size () > maxChain . size ()) maxChain = x ; for ( Pair pair : maxChain ) cout < < '(' < < pair . a < < ' ' < < pair . b < < ') ' ; } // Driver Function int main () { Pair a [] = {{ 5 29 } { 39 40 } { 15 28 } { 27 40 } { 50 90 }}; int n = sizeof ( a ) / sizeof ( a [ 0 ]); vector < Pair > arr ( a a + n ); maxChainLength ( arr ); return 0 ; }
Java // Java program to implement the approach import java.util.ArrayList ; import java.util.Collections ; import java.util.List ; // User Defined Pair Class class Pair { int a ; int b ; } class GFG { // Custom comparison function public static int compare ( Pair x Pair y ) { return x . a - ( y . a ); } public static void maxChainLength ( List < Pair > arr ) { // Sort by start time Collections . sort ( arr Main :: compare ); // L[i] stores maximum length of chain of // arr[0..i] that ends with arr[i]. List < List < Pair >> L = new ArrayList <> (); // L[0] is equal to arr[0] List < Pair > l0 = new ArrayList <> (); l0 . add ( arr . get ( 0 )); L . add ( l0 ); for ( int i = 0 ; i < arr . size () - 1 ; i ++ ) { L . add ( new ArrayList <> ()); } // start from index 1 for ( int i = 1 ; i < arr . size (); i ++ ) { // for every j less than i for ( int j = 0 ; j < i ; j ++ ) { // L[i] = {Max(L[j])} + arr[i] // where j < i and arr[j].b < arr[i].a if ( arr . get ( j ). b < arr . get ( i ). a && L . get ( j ). size () > L . get ( i ). size ()) L . set ( i L . get ( j )); } L . get ( i ). add ( arr . get ( i )); } // print max length vector List < Pair > maxChain = new ArrayList <> (); for ( List < Pair > x : L ) if ( x . size () > maxChain . size ()) maxChain = x ; for ( Pair pair : maxChain ) System . out . println ( '(' + pair . a + ' ' + pair . b + ') ' ); } // Driver Code public static void main ( String [] args ) { Pair [] a = { new Pair () {{ a = 5 ; b = 29 ;}} new Pair () {{ a = 39 ; b = 40 ;}} new Pair () {{ a = 15 ; b = 28 ;}} new Pair () {{ a = 27 ; b = 40 ;}} new Pair () {{ a = 50 ; b = 90 ;}}}; int n = a . length ; List < Pair > arr = new ArrayList <> (); for ( Pair anA : a ) { arr . add ( anA ); } // Function call maxChainLength ( arr ); } } // This code is contributed by phasing17
Python3 # Dynamic Programming solution to construct # Maximum Length Chain of Pairs class Pair : def __init__ ( self a b ): self . a = a self . b = b def __lt__ ( self other ): return self . a < other . a def maxChainLength ( arr ): # Function to construct # Maximum Length Chain of Pairs # Sort by start time arr . sort () # L[i] stores maximum length of chain of # arr[0..i] that ends with arr[i]. L = [[] for x in range ( len ( arr ))] # L[0] is equal to arr[0] L [ 0 ] . append ( arr [ 0 ]) # start from index 1 for i in range ( 1 len ( arr )): # for every j less than i for j in range ( i ): # L[i] = {Max(L[j])} + arr[i] # where j < i and arr[j].b < arr[i].a if ( arr [ j ] . b < arr [ i ] . a and len ( L [ j ]) > len ( L [ i ])): L [ i ] = L [ j ] L [ i ] . append ( arr [ i ]) # print max length vector maxChain = [] for x in L : if len ( x ) > len ( maxChain ): maxChain = x for pair in maxChain : print ( '( {a} {b} )' . format ( a = pair . a b = pair . b ) end = ' ' ) print () # Driver Code if __name__ == '__main__' : arr = [ Pair ( 5 29 ) Pair ( 39 40 ) Pair ( 15 28 ) Pair ( 27 40 ) Pair ( 50 90 )] n = len ( arr ) maxChainLength ( arr ) # This code is contributed # by vibhu4agarwal
C# using System ; using System.Collections.Generic ; public class Pair { public int a ; public int b ; } public class Program { public static int Compare ( Pair x Pair y ) { return x . a - ( y . a ); } public static void MaxChainLength ( List < Pair > arr ) { // Sort by start time arr . Sort ( Compare ); // L[i] stores maximum length of chain of // arr[0..i] that ends with arr[i]. List < List < Pair >> L = new List < List < Pair >> (); // L[0] is equal to arr[0] L . Add ( new List < Pair > { arr [ 0 ] }); for ( int i = 0 ; i < arr . Count - 1 ; i ++ ) L . Add ( new List < Pair > ()); // start from index 1 for ( int i = 1 ; i < arr . Count ; i ++ ) { // for every j less than i for ( int j = 0 ; j < i ; j ++ ) { // L[i] = {Max(L[j])} + arr[i] // where j < i and arr[j].b < arr[i].a if ( arr [ j ]. b < arr [ i ]. a && L [ j ]. Count > L [ i ]. Count ) L [ i ] = L [ j ]; } L [ i ]. Add ( arr [ i ]); } // print max length vector List < Pair > maxChain = new List < Pair > (); foreach ( List < Pair > x in L ) if ( x . Count > maxChain . Count ) maxChain = x ; foreach ( Pair pair in maxChain ) Console . WriteLine ( '(' + pair . a + ' ' + pair . b + ') ' ); } public static void Main () { Pair [] a = { new Pair () { a = 5 b = 29 } new Pair () { a = 39 b = 40 } new Pair () { a = 15 b = 28 } new Pair () { a = 27 b = 40 } new Pair () { a = 50 b = 90 } }; int n = a . Length ; List < Pair > arr = new List < Pair > ( a ); MaxChainLength ( arr ); } }
JavaScript < script > // Dynamic Programming solution to construct // Maximum Length Chain of Pairs class Pair { constructor ( a b ){ this . a = a this . b = b } } function maxChainLength ( arr ){ // Function to construct // Maximum Length Chain of Pairs // Sort by start time arr . sort (( c d ) => c . a - d . a ) // L[i] stores maximum length of chain of // arr[0..i] that ends with arr[i]. let L = new Array ( arr . length ). fill ( 0 ). map (()=> new Array ()) // L[0] is equal to arr[0] L [ 0 ]. push ( arr [ 0 ]) // start from index 1 for ( let i = 1 ; i < arr . length ; i ++ ){ // for every j less than i for ( let j = 0 ; j < i ; j ++ ){ // L[i] = {Max(L[j])} + arr[i] // where j < i and arr[j].b < arr[i].a if ( arr [ j ]. b < arr [ i ]. a && L [ j ]. length > L [ i ]. length ) L [ i ] = L [ j ] } L [ i ]. push ( arr [ i ]) } // print max length vector let maxChain = [] for ( let x of L ){ if ( x . length > maxChain . length ) maxChain = x } for ( let pair of maxChain ) document . write ( `( ${ pair . a } ${ pair . b } ) ` ) document . write ( ' ' ) } // driver code let arr = [ new Pair ( 5 29 ) new Pair ( 39 40 ) new Pair ( 15 28 ) new Pair ( 27 40 ) new Pair ( 50 90 )] let n = arr . length maxChainLength ( arr ) /// This code is contributed by shinjanpatra < /script>
Lähtö:
(5 29) (39 40) (50 90)
Aika monimutkaisuus yllä olevasta Dynaamisen ohjelmoinnin ratkaisusta on O(n 2 ) jossa n on parien lukumäärä. Aputila ohjelman käyttämä arvo on O(n 2 ).
