动态规划——0-1背包问题

/**
* @brief 0_1_Knapsack  dynamic programming
* @author An
* @data  2013.8.28
**/

/**
* @problem
* @0-1背包问题：
/*    给定n种物品和一个背包, 物品i的重量为wi，其价值为vi, 背包的容量为c,
/*    应如何选择装入背包的物品，使得装入背包中的物品的总价值最大?
/*    注：在选择装入背包的物品时，对物品i只有两种选择，
/*        即装入或不装入背包。不能将物品i装入多次，也
/*        不能只装入部分的物品i。
/*
/* 1. 0-1背包问题的形式化描述：
/*    给定c>0, wi>0, vi>0, 0<=i<=n，要求找到一个n元的
/*    0-1向量(x1, x2, ..., xn), 使得：
/*            max sum_{i=1 to n} (vi*xi),且满足如下约束：
/*        (1) sum_{i=1 to n} (wi*xi) <= c
/*        (2) xi∈{0, 1}, 1<=i<=n
/*
* @
**/

#include <iostream>
#define max( x, y ) ( x >= y ? x : y )
using namespace std;

void Knapsack( int *weight, double *value, int capacity, int n, bool *res, double **V );

int main()
{
int w[] = { 2, 2, 6, 5, 4 };
int v[] = { 6, 3, 5, 4, 6 };
int n = 5;
int *weight = new int
;
double *value = new double
;
for ( int i = 0; i < n; ++i )
{
weight[i] = w[i];
value[i] = v[i];
}
int capacity = 10;

// distribute memory
bool *res = new bool
;
double **V = new double*[n + 1];
for ( int i = 0; i <= n; ++i )
{
V[i] = new double[capacity + 1];
}

Knapsack( weight, value, capacity, n, res, V );

cout << "the max value is: " << V
[capacity] << endl;
cout << "the items in the knapsack is: ";
for ( int i = 0; i != n; ++i )
{
if ( res[i] )
{
cout << i + 1 << ", ";
}
}
cout << endl;

return 0;
}

void Knapsack( int *weight, double *value, int capacity, int n, bool *res, double **V )
{

// initialize 0 row and 0 column
for ( int i = 0; i <= n; ++i )
{
V[i][0] = 0;
}
for ( int j = 0; j <= capacity; ++j )
{
V[0][j] = 0;
}

// calculate V[][]
for ( int i = 1; i <= n; ++i )
{
for ( int j = 1; j <=capacity; ++j )
{
if ( j < weight[i - 1] )
{
V[i][j] = V[i - 1][j];
}
else
{
V[i][j] = max( V[i - 1][j], V[i - 1][j - weight[i - 1]] + value[i - 1] );
}
}
}

int j = capacity;
for ( int i = n; i > 0; --i )
{
if ( V[i][j] > V[i - 1][j] )
{
res[i - 1] = true;
j -= weight[i - 1];
}
else
{
res[i - 1] = false;
}
}
}