Dynamic planning Three: 0-1 knapsack problem

1. Description of the problem:

A certain object and a backpack, the weight of the object I is the value of the WI for the VI, the capacity of the backpack is C, how to put the maximum value of the backpack? The problem can be described as:

　　

　　

2. Problem Analysis:

1) Optimal sub-structure:

　　

where J=c-wiyi

2) Recursive relationship: Set the optimal value of M (I,j), J for the optimal capacity, I for optional items, by the optimal substructure properties can be established recursive:

　　

　　

3. Algorithm Description:(not optimized)

1#include <iostream>2 using namespacestd;3 intc[Ten][ -];4 intKnap (intMintN) {5     intw[Ten];6     intv[Ten];7cout <<"input Object Weight W and value v:\n";8      for(inti =1; I <= N; i++){9CIN >> W[i] >>V[i];Ten     } One      for(inti =1; I <= N; i++) A     { -          for(intj =1; J <= M; J + +)//the capacity of the backpack is tested individually -         { the             if(W[i] <=j) -             { -                 if(C[i-1][J] < V[i] + c[i-1][j-W[i]]) -C[I][J] = V[i] + c[i-1][j-W[i]]; +                 Else -C[I][J] = c[i-1][j]; +             } A             Else atC[I][J] = c[i-1][j]; -         } -     } -     returnC[n][m]; - } -  in intMain () { -     intN, M; tocout <<"number of input objects N and backpack capacity m:\n"; +CIN >> N >>m; -cout << Knap (n, m) <<Endl; the      for(inti =0; I <= N; i++) *     { $          for(intj =0; J <= M; J + +)Panax Notoginsengcout<<C[i][j]; -cout<<Endl; the     } +     return 0; A}

　　

Dynamic planning Three: 0-1 knapsack problem