Java程序0-1背包问题
递归解决方案
/* A Naive recursive implementation of 0-1 Knapsack problem */
class Knapsack {
// A utility function that returns maximum of two integers
static int max(int a, int b) { return (a > b) ? a : b; }
// Returns the maximum value that can
// be put in a knapsack of capacity W
static int knapSack(int W, int wt[], int val[], int n)
{
// Base Case
if (n == 0 || W == 0)
return 0;
// If weight of the nth item is more
// than Knapsack capacity W, then
// this item cannot be included in the optimal solution
if (wt[n - 1] > W)
return knapSack(W, wt, val, n - 1);
// Return the maximum of two cases:
// (1) nth item included
// (2) not included
else
return max(val[n - 1] + knapSack(W - wt[n - 1], wt, val, n - 1),
knapSack(W, wt, val, n - 1));
}
// Driver program to test above function
public static void main(String args[])
{
int val[] = new int[] { 60, 100, 120 };
int wt[] = new int[] { 10, 20, 30 };
int W = 50;
int n = val.length;
System.out.println(knapSack(W, wt, val, n));
}
}
/*This code is contributed by Rajat Mishra */
输出:
220
动态规划解决方案
// A Dynamic Programming based solution for 0-1 Knapsack problem
class Knapsack {
// A utility function that returns maximum of two integers
static int max(int a, int b)
{ return (a > b) ? a : b; }
// Returns the maximum value that can be put in a knapsack
// of capacity W
static int knapSack(int W, int wt[], int val[], int n)
{
int i, w;
int K[][] = new int[n + 1][W + 1];
// Build table K[][] in bottom up manner
for (i = 0; i<= n; i++) {
for (w = 0; w<= W; w++) {
if (i == 0 || w == 0)
K[i][w] = 0;
else if (wt[i - 1]<= w)
K[i][w] = max(val[i - 1] + K[i - 1][w - wt[i - 1]], K[i - 1][w]);
else
K[i][w] = K[i - 1][w];
}
}
return K[n][W];
}
// Driver program to test above function
public static void main(String args[])
{
int val[] = new int[] { 60, 100, 120 };
int wt[] = new int[] { 10, 20, 30 };
int W = 50;
int n = val.length;
System.out.println(knapSack(W, wt, val, n));
}
}
/*This code is contributed by Rajat Mishra */
输出:
220
请参阅有关动态规划的完整文章 |第 10 套(0-1 背包问题)了解更多详情!