切割棒的Java程序| DP-13
给定一根长度为 n 英寸的杆和一个价格数组，其中包含所有尺寸小于 n 的块的价格。确定通过切割杆并出售碎片可获得的最大值。例如，如果杆的长度为 8 并且不同部分的值如下给出，则最大可获得值是 22（通过切割长度为 2 和 6 的两根）
length | 1 2 3 4 5 6 7 8
--------------------------------------------
price | 1 5 8 9 10 17 17 20如果价格如下，那么最大可获得值为24（通过切割8块长度1）
length | 1 2 3 4 5 6 7 8
--------------------------------------------
price | 3 5 8 9 10 17 17 20以下是棒切割问题的简单递归实现。实现简单地遵循上面提到的递归结构。
Java
// // A Naive recursive solution for Rod cutting problem
class RodCutting {
/* Returns the best obtainable price for a rod of length
n and price[] as prices of different pieces */
static int cutRod(int price[], int n)
{
if (n <= 0)
return 0;
int max_val = Integer.MIN_VALUE;
// Recursively cut the rod in different pieces and
// compare different configurations
for (int i = 0; i < n; i++)
max_val = Math.max(max_val,
price[i] + cutRod(price, n - i - 1));
return max_val;
}
/* Driver program to test above functions */
public static void main(String args[])
{
int arr[] = new int[] { 1, 5, 8, 9, 10, 17, 17, 20 };
int size = arr.length;
System.out.println("Maximum Obtainable Value is " + cutRod(arr, size));
}
}
/* This code is contributed by Rajat Mishra */Java
// A Dynamic Programming solution for Rod cutting problem
class RodCutting {
/* Returns the best obtainable price for a rod of
length n and price[] as prices of different pieces */
static int cutRod(int price[], int n)
{
int val[] = new int[n + 1];
val[0] = 0;
// Build the table val[] in bottom up manner and return
// the last entry from the table
for (int i = 1; i <= n; i++) {
int max_val = Integer.MIN_VALUE;
for (int j = 0; j < i; j++)
max_val = Math.max(max_val,
price[j] + val[i - j - 1]);
val[i] = max_val;
}
return val[n];
}
/* Driver program to test above functions */
public static void main(String args[])
{
int arr[] = new int[] { 1, 5, 8, 9, 10, 17, 17, 20 };
int size = arr.length;
System.out.println("Maximum Obtainable Value is " + cutRod(arr, size));
}
}
/* This code is contributed by Rajat Mishra */输出：
Maximum Obtainable Value is 22考虑到上述实现，以下是长度为 4 的 Rod 的递归树。
cR() ---> cutRod()
cR(4)
/ /
/ /
cR(3) cR(2) cR(1) cR(0)
/ | / |
/ | / |
cR(2) cR(1) cR(0) cR(1) cR(0) cR(0)
/ | |
/ | |
cR(1) cR(0) cR(0) cR(0)
/
/
CR(0)在上面的部分递归树中，cR(2) 被求解了两次。我们可以看到有很多子问题被一次又一次地解决。由于再次调用相同的子问题，此问题具有重叠子问题属性。因此，Rod Cutting 问题具有动态规划问题的两个属性（参见 this 和 this）。与其他典型的动态规划（DP）问题一样，通过以自下而上的方式构造临时数组 val[] 可以避免重复计算相同的子问题。
Java
// A Dynamic Programming solution for Rod cutting problem
class RodCutting {
/* Returns the best obtainable price for a rod of
length n and price[] as prices of different pieces */
static int cutRod(int price[], int n)
{
int val[] = new int[n + 1];
val[0] = 0;
// Build the table val[] in bottom up manner and return
// the last entry from the table
for (int i = 1; i <= n; i++) {
int max_val = Integer.MIN_VALUE;
for (int j = 0; j < i; j++)
max_val = Math.max(max_val,
price[j] + val[i - j - 1]);
val[i] = max_val;
}
return val[n];
}
/* Driver program to test above functions */
public static void main(String args[])
{
int arr[] = new int[] { 1, 5, 8, 9, 10, 17, 17, 20 };
int size = arr.length;
System.out.println("Maximum Obtainable Value is " + cutRod(arr, size));
}
}
/* This code is contributed by Rajat Mishra */
输出：
Maximum Obtainable Value is 22请参阅有关切割棒的完整文章 | DP-13 了解更多详情！