Svertinis darbo planavimas | 2 rinkinys (naudojant LIS)
Duota N darbų, kur kiekvienas darbas vaizduojamas trimis jo elementais.
1. Pradžios laikas
2. Baigimo laikas
3. Su pelnu ar verte
Raskite maksimalaus pelno darbų poaibį, kad poaibyje nebūtų dviejų darbų.
Pavyzdžiai:
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.Į ankstesnis pranešimas, kurį aptarėme apie svertinio darbo planavimo problemą. Aptarėme DP sprendimą, kuriame iš esmės įtraukiame arba neįtraukiame esamo darbo. Šiame įraše aptariamas kitas įdomus DP sprendimas, kuriame taip pat spausdiname darbus. Ši problema yra standarto variantas Ilgiausiai didėjanti seka (LIS) problema. Mums reikia šiek tiek pakeisti LIS problemos dinaminio programavimo sprendimą.
Pirmiausia turime surūšiuoti darbus pagal pradžios laiką. Tegul darbas[0..n-1] yra užduočių masyvas po rūšiavimo. Vektorių L apibrėžiame taip, kad pats L[i] yra vektorius, kuriame saugomas užduoties [0..i] svertinis darbo planavimas, kuris baigiasi užduotimi [i]. Todėl indeksui i L[i] galima rekursyviai parašyti kaip -
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
Pavyzdžiui, apsvarstykite poras {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}Mes pasirenkame vektorių su didžiausiu pelnu. Šiuo atveju L[1].
Žemiau yra aukščiau pateiktos idėjos įgyvendinimas -
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 ); } }
Išvestis(1 2 50) (2 100 200)
Mes galime toliau optimizuoti aukščiau pateiktą DP sprendimą pašalindami funkciją findSum (). Vietoj to galime išlaikyti kitą vektorių / masyvą, kad išsaugotume didžiausio galimo pelno sumą iki darbo i.Laiko sudėtingumas dinaminio programavimo sprendimas yra O(n 2 ) kur n yra darbų skaičius.
Pagalbinė erdvė programos naudojamas O (n 2 ).
