[C++11] Algorithm [Exhaustive] output knapsack problem of all the satisfied solutions

On the question of knapsack problem, the previous statement is prepared, here only to discuss the realization

Input:

N

Ca

W_1 v_1

W_2 v_2

...

W_n V_n

where n is the total number of items, CA is backpack size, w_n is the weight of nth item, V_n is the value of nth item

Output:

V_1 x

V_2 x

V_3 x

...

Where V_n is the value of the backpack at the time of X, X is a sequence of 0, 1, indicating whether to put the backpack

Example: 1001 means the first and last item is put into the backpack, the middle two items are not put in

Requires writing a program that outputs all the solutions that can be satisfied.

The idea is simple, that is, poor lifting. Poor to lift every situation.

The pseudo code is as follows:

f () {  if  (nothing to put in the backpack) {    output   else  {    put in Backpack    f ()    not put in backpack    f () Restitution  }}

In this way, a binary tree is constructed that can output each of the optional solutions.

My code is as follows:

1#include <iostream>2#include <functional>3 4 structPack {5 unsigned cnt;6unsigned *w;//Weights7unsigned *v;//Values8unsigned *x;//put in or not9unsigned CA;//capacityTen    One Pack (unsigned items_cnt): CNT (items_cnt) { AW =Newunsigned [items_cnt]; -v =Newunsigned [items_cnt]; -x =Newunsigned [items_cnt]; the      -      for(inti =0; i < items_cnt; ++i) { -W[i] = V[i] = X[i] =-1; -     } +   } -    +~Pack () { A     Delete[] W; at     Delete[] v; -     Delete[] x; -   } - }; -  -  in intMain () { - unsigned C; to   +Std::cin >>C; -    the Pack P (c); *    $Std::cin >>p.ca;Panax Notoginseng    -    for(inti =0; i < p.cnt; ++i) { theStd::cin >> P.w[i] >>P.v[i]; +   } A   the   +   -   $Std::function<unsigned () > Totval = [&]() { $unsigned cnt =0; -      -      for(inti =0; i < p.ca; ++i) { the       if(P.x[i] = =1) CNT + =P.v[i]; -     }Wuyi      the     returnCNT; -   }; Wu    -Std::function<unsigned () > TOTWT = [&]() { Aboutunsigned cnt =0; $      -      for(inti =0; i < p.ca; ++i) { -       if(P.x[i] = =1) CNT + =P.w[i]; -     } A      +     returnCNT; the   }; -    $std::function<void() > Loop = [&]() { theunsigned no =-1; the      for(inti =0; i < p.cnt; ++i) { the       if(P.x[i] = =-1) { theNo =i; -          Break; in       } the     } the      About     if(No = =-1) { theStd::cout << totval () <<' '; the        for(inti =0; i < p.cnt; ++i) { theStd::cout <<P.x[i]; +       } -Std::cout <<Std::endl; the}Else {BayiP.x[no] =0; the loop (); the        -P.x[no] =1; -        the       if(TOTWT () <=p.ca) { the loop (); the       } the        -P.x[no] =-1; the     } the   }; the   94 loop (); the   the   return 0; the}

Test data:

5  the 9 8 1 2 3 8 1 0 3 9

Output:

0 000009 000010 000109 000118 00100 - 001018 00110 - 001112 01000 One 010012 01010 One 01011Ten 01100 + 01101Ten 01110 + 011118 10000 - 100018 10010 - 10011 - 10100 - 10101 - 10110Ten 11000 + 11001Ten 11010 + 11011 - 11100 - 11110

The optimal solution can be found, 10101.

[C++11] Algorithm [Exhaustive] output knapsack problem of all the satisfied solutions