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[n];double *value = new double[n];for ( int i = 0; i < n; ++i ){weight[i] = w[i];value[i] = v[i];}int capacity = 10;// distribute memorybool *res = new bool[n];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[n][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 columnfor ( 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;}}}