Programació de treball ponderada | Conjunt 2 (utilitzant LIS)
Donats N llocs de treball on cada lloc de treball es representa seguint-ne tres elements.
1. Hora d'inici
2. Hora d'acabament
3. Benefici o valor associat
Trobeu el subconjunt de beneficis màxims de llocs de treball de manera que no es superposin dos llocs de treball del subconjunt.
Exemples:
Input:
Number of Jobs n = 4
Job Details {Start Time Finish Time Profit}
Job 1: {1 2 50}
Job 2: {3 5 20}
Job 3: {6 19 100}
Job 4: {2 100 200}
Output:
Job 1: {1 2 50}
Job 4: {2 100 200}
Explanation: We can get the maximum profit by
scheduling jobs 1 and 4 and maximum profit is 250.En anterior publicació que hem comentat sobre el problema de la programació de treballs ponderats. Hem parlat d'una solució de DP on bàsicament incloem o excloem la feina actual. En aquest post es parla d'una altra solució DP interessant on també imprimim els treballs. Aquest problema és una variació de l'estàndard Subseqüència creixent més llarga (LIS) problema. Necessitem un lleuger canvi en la solució de programació dinàmica del problema LIS.
Primer hem d'ordenar les feines segons l'hora d'inici. Sigui treball[0..n-1] la matriu de treballs després de l'ordenació. Definim el vector L de manera que L[i] és en si mateix un vector que emmagatzema la programació ponderada de treballs del treball[0..i] que acaba amb el treball[i]. Per tant, per a un índex i L[i] es pot escriure recursivament com -
L[0] = {job[0]}
L[i] = {MaxSum(L[j])} + job[i] where j < i and job[j].finish <= job[i].start
= job[i] if there is no such j
Per exemple, considereu les parelles {3 10 20} {1 2 50} {6 19 100} {2 100 200}After sorting we get
{1 2 50} {2 100 200} {3 10 20} {6 19 100}
Therefore
L[0]: {1 2 50}
L[1]: {1 2 50} {2 100 200}
L[2]: {1 2 50} {3 10 20}
L[3]: {1 2 50} {6 19 100}Escollim el vector amb més beneficis. En aquest cas L[1].
A continuació es mostra la implementació de la idea anterior:
C++Java// C++ program for weighted job scheduling using LIS #include#include #include using namespace std ; // A job has start time finish time and profit. struct Job { int start finish profit ; }; // Utility function to calculate sum of all vector // elements int findSum ( vector < Job > arr ) { int sum = 0 ; for ( int i = 0 ; i < arr . size (); i ++ ) sum += arr [ i ]. profit ; return sum ; } // comparator function for sort function int compare ( Job x Job y ) { return x . start < y . start ; } // The main function that finds the maximum possible // profit from given array of jobs void findMaxProfit ( vector < Job > & arr ) { // Sort arr[] by start time. sort ( arr . begin () arr . end () compare ); // L[i] stores Weighted Job Scheduling of // job[0..i] that ends with job[i] vector < vector < Job >> 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] = {MaxSum(L[j])} + arr[i] where j < i // and arr[j].finish <= arr[i].start if (( arr [ j ]. finish <= arr [ i ]. start ) && ( findSum ( L [ j ]) > findSum ( L [ i ]))) L [ i ] = L [ j ]; } L [ i ]. push_back ( arr [ i ]); } vector < Job > maxChain ; // find one with max profit for ( int i = 0 ; i < L . size (); i ++ ) if ( findSum ( L [ i ]) > findSum ( maxChain )) maxChain = L [ i ]; for ( int i = 0 ; i < maxChain . size (); i ++ ) cout < < '(' < < maxChain [ i ]. start < < ' ' < < maxChain [ i ]. finish < < ' ' < < maxChain [ i ]. profit < < ') ' ; } // Driver Function int main () { Job a [] = { { 3 10 20 } { 1 2 50 } { 6 19 100 } { 2 100 200 } }; int n = sizeof ( a ) / sizeof ( a [ 0 ]); vector < Job > arr ( a a + n ); findMaxProfit ( arr ); return 0 ; } Python// Java program for weighted job // scheduling using LIS import java.util.ArrayList ; import java.util.Arrays ; import java.util.Collections ; import java.util.Comparator ; class Graph { // A job has start time finish time // and profit. static class Job { int start finish profit ; public Job ( int start int finish int profit ) { this . start = start ; this . finish = finish ; this . profit = profit ; } }; // Utility function to calculate sum of all // ArrayList elements static int findSum ( ArrayList < Job > arr ) { int sum = 0 ; for ( int i = 0 ; i < arr . size (); i ++ ) sum += arr . get ( i ). profit ; return sum ; } // The main function that finds the maximum // possible profit from given array of jobs static void findMaxProfit ( ArrayList < Job > arr ) { // Sort arr[] by start time. Collections . sort ( arr new Comparator < Job > () { @Override public int compare ( Job x Job y ) { return x . start - y . start ; } }); // sort(arr.begin() arr.end() compare); // L[i] stores Weighted Job Scheduling of // job[0..i] that ends with job[i] ArrayList < ArrayList < Job >> L = new ArrayList <> (); for ( int i = 0 ; i < arr . size (); i ++ ) { L . add ( new ArrayList <> ()); } // L[0] is equal to arr[0] L . get ( 0 ). add ( arr . get ( 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] = {MaxSum(L[j])} + arr[i] where j < i // and arr[j].finish <= arr[i].start if (( arr . get ( j ). finish <= arr . get ( i ). start ) && ( findSum ( L . get ( j )) > findSum ( L . get ( i )))) { ArrayList < Job > copied = new ArrayList <> ( L . get ( j )); L . set ( i copied ); } } L . get ( i ). add ( arr . get ( i )); } ArrayList < Job > maxChain = new ArrayList <> (); // Find one with max profit for ( int i = 0 ; i < L . size (); i ++ ) if ( findSum ( L . get ( i )) > findSum ( maxChain )) maxChain = L . get ( i ); for ( int i = 0 ; i < maxChain . size (); i ++ ) { System . out . printf ( '(%d %d %d)n' maxChain . get ( i ). start maxChain . get ( i ). finish maxChain . get ( i ). profit ); } } // Driver code public static void main ( String [] args ) { Job [] a = { new Job ( 3 10 20 ) new Job ( 1 2 50 ) new Job ( 6 19 100 ) new Job ( 2 100 200 ) }; ArrayList < Job > arr = new ArrayList <> ( Arrays . asList ( a )); findMaxProfit ( arr ); } } // This code is contributed by sanjeev2552C## Python program for weighted job scheduling using LIS import sys # A job has start time finish time and profit. class Job : def __init__ ( self start finish profit ): self . start = start self . finish = finish self . profit = profit # Utility function to calculate sum of all vector elements def findSum ( arr ): sum = 0 for i in range ( len ( arr )): sum += arr [ i ] . profit return sum # comparator function for sort function def compare ( x y ): if x . start < y . start : return - 1 elif x . start == y . start : return 0 else : return 1 # The main function that finds the maximum possible profit from given array of jobs def findMaxProfit ( arr ): # Sort arr[] by start time. arr . sort ( key = lambda x : x . start ) # L[i] stores Weighted Job Scheduling of job[0..i] that ends with job[i] L = [[] for _ 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] = {MaxSum(L[j])} + arr[i] where j < i # and arr[j].finish <= arr[i].start if arr [ j ] . finish <= arr [ i ] . start and findSum ( L [ j ]) > findSum ( L [ i ]): L [ i ] = L [ j ][:] L [ i ] . append ( arr [ i ]) maxChain = [] # find one with max profit for i in range ( len ( L )): if findSum ( L [ i ]) > findSum ( maxChain ): maxChain = L [ i ] for i in range ( len ( maxChain )): print ( '( {} {} {} )' . format ( maxChain [ i ] . start maxChain [ i ] . finish maxChain [ i ] . profit ) end = ' ' ) # Driver Function if __name__ == '__main__' : a = [ Job ( 3 10 20 ) Job ( 1 2 50 ) Job ( 6 19 100 ) Job ( 2 100 200 )] findMaxProfit ( a )JavaScriptusing System ; using System.Collections.Generic ; using System.Linq ; public class Graph { // A job has start time finish time // and profit. public class Job { public int start finish profit ; public Job ( int start int finish int profit ) { this . start = start ; this . finish = finish ; this . profit = profit ; } }; // Utility function to calculate sum of all // ArrayList elements public static int FindSum ( List < Job > arr ) { int sum = 0 ; for ( int i = 0 ; i < arr . Count ; i ++ ) sum += arr . ElementAt ( i ). profit ; return sum ; } // The main function that finds the maximum // possible profit from given array of jobs public static void FindMaxProfit ( List < Job > arr ) { // Sort arr[] by start time. arr . Sort (( x y ) => x . start . CompareTo ( y . start )); // L[i] stores Weighted Job Scheduling of // job[0..i] that ends with job[i] List < List < Job >> L = new List < List < Job >> (); for ( int i = 0 ; i < arr . Count ; i ++ ) { L . Add ( new List < Job > ()); } // L[0] is equal to arr[0] L [ 0 ]. Add ( arr [ 0 ]); // 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] = {MaxSum(L[j])} + arr[i] where j < i // and arr[j].finish <= arr[i].start if (( arr [ j ]. finish <= arr [ i ]. start ) && ( FindSum ( L [ j ]) > FindSum ( L [ i ]))) { List < Job > copied = new List < Job > ( L [ j ]); L [ i ] = copied ; } } L [ i ]. Add ( arr [ i ]); } List < Job > maxChain = new List < Job > (); // Find one with max profit for ( int i = 0 ; i < L . Count ; i ++ ) if ( FindSum ( L [ i ]) > FindSum ( maxChain )) maxChain = L [ i ]; for ( int i = 0 ; i < maxChain . Count ; i ++ ) { Console . WriteLine ( '({0} {1} {2})' maxChain [ i ]. start maxChain [ i ]. finish maxChain [ i ]. profit ); } } // Driver code public static void Main ( String [] args ) { Job [] a = { new Job ( 3 10 20 ) new Job ( 1 2 50 ) new Job ( 6 19 100 ) new Job ( 2 100 200 ) }; List < Job > arr = new List < Job > ( a ); FindMaxProfit ( arr ); } }
Sortida(1 2 50) (2 100 200)
Podem optimitzar encara més la solució DP anterior eliminant la funció findSum(). En lloc d'això, podem mantenir un altre vector/matriu per emmagatzemar la suma del màxim benefici possible fins al treball i.Complexitat temporal La solució de programació dinàmica anterior és O(n 2 ) on n és el nombre de llocs de treball.
Espai auxiliar utilitzat pel programa és O(n 2 ).
