Imprimați lanțul de perechi cu lungime maximă
Vi se dau n perechi de numere. În fiecare pereche, primul număr este întotdeauna mai mic decât al doilea număr. O pereche (c d) poate urma o altă pereche (a b) dacă 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. Exemple:
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) În anterior post am discutat despre problema Lanțului de perechi cu lungime maximă. Cu toate acestea, postarea a acoperit doar codul legat de găsirea lungimii lanțului de dimensiune maximă, dar nu și de construcția lanțului de dimensiune maximă. În această postare vom discuta despre cum să construim propriul lanț de perechi cu lungime maximă. Ideea este de a sorta mai întâi perechile date în ordinea crescătoare a primului lor element. Fie arr[0..n-1] tabloul de intrare de perechi după sortare. Definim vectorul L astfel încât L[i] este el însuși un vector care stochează Lanțul de Lungime Maximă de Perechi de arr[0..i] care se termină cu arr[i]. Prin urmare, pentru un indice i L[i] poate fi scris recursiv ca -
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 De exemplu pentru (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)
Vă rugăm să rețineți că sortarea perechilor se face, deoarece trebuie să găsim lungimea maximă a perechilor, iar ordinea nu contează aici. Dacă nu sortăm, vom obține perechi în ordine crescătoare, dar nu vor fi perechi maxime posibile. Mai jos este implementarea ideii de mai sus -
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>
Ieșire:
(5 29) (39 40) (50 90)
Complexitatea timpului soluția de programare dinamică de mai sus este O(n 2 ) unde n este numărul de perechi. Spațiu auxiliar folosit de program este O(n 2 ).
