满足给定重量的袋子的最低成本

给您一个大小为W kg的袋子，并为您提供数组cost []中重量不同的橙子的包装成本，其中cost [i]基本上是‘ i’kg橙子包装的成本。其中cost [i] = -1表示没有‘i’公斤橙色包装
找出购买总重为W kg的橘子的最低总成本，如果不可能购买总重为W kg的橘子，则打印-1。可以假设所有可用的数据包类型都有无限的供应。
注意：数组从索引1开始。

例子：

Input  : W = 5, cost[] = {20, 10, 4, 50, 100}
Output : 14
We can choose two oranges to minimize cost. First 
orange of 2Kg and cost 10. Second orange of 3Kg
and cost 4. 

Input  : W = 5, cost[] = {1, 10, 4, 50, 100}
Output : 5
We can choose five oranges of weight 1 kg.

Input  : W = 5, cost[] = {1, 2, 3, 4, 5}
Output : 5
Costs of 1, 2, 3, 4 and 5 kg packets are 1, 2, 3, 
4 and 5 Rs respectively. 
We choose packet of 5kg having cost 5 for minimum
cost to get 5Kg oranges.

Input  : W = 5, cost[] = {-1, -1, 4, 5, -1}
Output : -1
Packets of size 1, 2 and 5 kg are unavailable
because they have cost -1. Cost of 3 kg packet 
is 4 Rs and of 4 kg is 5 Rs. Here we have only 
weights 3 and 4 so by using these two we can  
not make exactly W kg weight, therefore answer 
is -1.

此问题可以简化为“无限制背包”。因此，在成本数组中，我们首先忽略那些不可用的数据包，即； cost是-1，然后遍历cost数组并创建两个数组val []来存储‘ i’kg橙包装的成本，而wt []来存储相应包装的重量。假设cost [i] = 50，则数据包的权重为i，成本为50。
算法 ：

• 创建矩阵min_cost [n + 1] [W + 1]，其中n是橙色的不同加权数据包的数量，W是袋子的最大容量。
• 用INF(无穷大)初始化第0行，用0初始化第0列。
• 现在填充矩阵
  • 如果wt [i-1]> j，则min_cost [i] [j] = min_cost [i-1] [j]；
  • 如果wt [i-1] <= j，则min_cost [i] [j] = min(min_cost [i-1] [j]，val [i-1] + min_cost [i] [j-wt [i-1 ]]);
• 如果min_cost [n] [W] == INF，则输出将为-1，因为这意味着我们无法使用这些权重来使制造权重W，否则输出将为min_cost [n] [W] 。

C++

// C++ program to find minimum cost to get exactly
// W Kg with given packets
#include
#define INF 1000000
using namespace std;
 
// cost[] initial cost array including unavailable packet
// W capacity of bag
int MinimumCost(int cost[], int n, int W)
{
    // val[] and wt[] arrays
    // val[] array to store cost of 'i' kg packet of orange
    // wt[] array weight of packet of orange
    vector val, wt;
 
    // traverse the original cost[] array and skip
    // unavailable packets and make val[] and wt[]
    // array. size variable tells the available number
    // of distinct weighted packets
    int size = 0;
    for (int i=0; ij means capacity of bag is
            // less then weight of item
            if (wt[i-1] > j)
                min_cost[i][j] = min_cost[i-1][j];
 
            // here we check we get minimum cost either
            // by including it or excluding it
            else
                min_cost[i][j] = min(min_cost[i-1][j],
                     min_cost[i][j-wt[i-1]] + val[i-1]);
        }
    }
 
    // exactly weight W can not be made by given weights
    return (min_cost[n][W]==INF)? -1: min_cost[n][W];
}
 
// Driver program to run the test case
int main()
{
    int cost[] = {1, 2, 3, 4, 5}, W = 5;
    int n = sizeof(cost)/sizeof(cost[0]);
 
    cout << MinimumCost(cost, n, W);
    return 0;
}

Java

// Java Code for Minimum cost to
// fill given weight in a bag
import java.util.*;
 
class GFG {
     
    // cost[] initial cost array including
    // unavailable packet W capacity of bag
    public static int MinimumCost(int cost[], int n,
                                             int W)
    {
        // val[] and wt[] arrays
        // val[] array to store cost of 'i' kg
        // packet of orange wt[] array weight of
        // packet of orange
        Vector val = new Vector();
        Vector wt = new Vector();
      
        // traverse the original cost[] array and skip
        // unavailable packets and make val[] and wt[]
        // array. size variable tells the available
        // number of distinct weighted packets
        int size = 0;
        for (int i = 0; i < n; i++)
        {
            if (cost[i] != -1)
            {
                val.add(cost[i]);
                wt.add(i + 1);
                size++;
            }
        }
      
        n = size;
        int min_cost[][] = new int[n+1][W+1];
      
        // fill 0th row with infinity
        for (int i = 0; i <= W; i++)
            min_cost[0][i] = Integer.MAX_VALUE;
      
        // fill 0'th column with 0
        for (int i = 1; i <= n; i++)
            min_cost[i][0] = 0;
      
        // now check for each weight one by one and
        // fill the matrix according to the condition
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= W; j++)
            {
                // wt[i-1]>j means capacity of bag is
                // less then weight of item
                if (wt.get(i-1) > j)
                    min_cost[i][j] = min_cost[i-1][j];
      
                // here we check we get minimum cost
                // either by including it or excluding
                // it
                else
                    min_cost[i][j] = Math.min(min_cost[i-1][j],
                                  min_cost[i][j-wt.get(i-1)] +
                                              val.get(i-1));
            }
        }
      
        // exactly weight W can not be made by
        // given weights
        return (min_cost[n][W] == Integer.MAX_VALUE)? -1:
                                        min_cost[n][W];
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
         int cost[] = {1, 2, 3, 4, 5}, W = 5;
            int n = cost.length;
          
        System.out.println(MinimumCost(cost, n, W));
    }
}
// This code is contributed by Arnav Kr. Mandal.

Python3

# Python program to find minimum cost to get exactly
# W Kg with given packets
 
INF = 1000000
 
# cost[] initial cost array including unavailable packet
# W capacity of bag
def MinimumCost(cost, n, W):
 
    # val[] and wt[] arrays
    # val[] array to store cost of 'i' kg packet of orange
    # wt[] array weight of packet of orange
    val = list()
    wt= list()
 
    # traverse the original cost[] array and skip
    # unavailable packets and make val[] and wt[]
    # array. size variable tells the available number
    # of distinct weighted packets.
    size = 0
    for i in range(n):
        if (cost[i] != -1):
            val.append(cost[i])
            wt.append(i+1)
            size += 1
 
    n = size
    min_cost = [[0 for i in range(W+1)] for j in range(n+1)]
 
    # fill 0th row with infinity
    for i in range(W+1):
        min_cost[0][i] = INF
 
    # fill 0th column with 0
    for i in range(1, n+1):
        min_cost[i][0] = 0
 
    # now check for each weight one by one and fill the
    # matrix according to the condition
    for i in range(1, n+1):
        for j in range(1, W+1):
            # wt[i-1]>j means capacity of bag is
            # less than weight of item
            if (wt[i-1] > j):
                min_cost[i][j] = min_cost[i-1][j]
 
            # here we check we get minimum cost either
            # by including it or excluding it
            else:
                min_cost[i][j] = min(min_cost[i-1][j],
                    min_cost[i][j-wt[i-1]] + val[i-1])
 
    # exactly weight W can not be made by given weights
    if(min_cost[n][W] == INF):
        return -1
    else:
        return min_cost[n][W]
 
# Driver program to run the test case
cost = [1, 2, 3, 4, 5]
W = 5
n = len(cost)
 
print(MinimumCost(cost, n, W))
 
# This code is contributed by Soumen Ghosh.

C#

// C# Code for Minimum cost to
// fill given weight in a bag
 
using System;
using System.Collections.Generic;
 
class GFG {
     
    // cost[] initial cost array including
    // unavailable packet W capacity of bag
    public static int MinimumCost(int []cost, int n,
                                            int W)
    {
        // val[] and wt[] arrays
        // val[] array to store cost of 'i' kg
        // packet of orange wt[] array weight of
        // packet of orange
        List val = new List();
        List wt = new List();
     
        // traverse the original cost[] array and skip
        // unavailable packets and make val[] and wt[]
        // array. size variable tells the available
        // number of distinct weighted packets
        int size = 0;
        for (int i = 0; i < n; i++)
        {
            if (cost[i] != -1)
            {
                val.Add(cost[i]);
                wt.Add(i + 1);
                size++;
            }
        }
     
        n = size;
        int [,]min_cost = new int[n+1,W+1];
     
        // fill 0th row with infinity
        for (int i = 0; i <= W; i++)
            min_cost[0,i] = int.MaxValue;
     
        // fill 0'th column with 0
        for (int i = 1; i <= n; i++)
            min_cost[i,0] = 0;
     
        // now check for each weight one by one and
        // fill the matrix according to the condition
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= W; j++)
            {
                // wt[i-1]>j means capacity of bag is
                // less then weight of item
                if (wt[i-1] > j)
                    min_cost[i,j] = min_cost[i-1,j];
     
                // here we check we get minimum cost
                // either by including it or excluding
                // it
                else
                    min_cost[i,j] = Math.Min(min_cost[i-1,j],
                                min_cost[i,j-wt[i-1]] + val[i-1]);
            }
        }
     
        // exactly weight W can not be made by
        // given weights
        return (min_cost[n,W] == int.MaxValue )? -1: min_cost[n,W];
    }
     
    /* Driver program to test above function */
    public static void Main()
    {
            int []cost = {1, 2, 3, 4, 5};
            int W = 5;
            int n = cost.Length;
         
            Console.WriteLine(MinimumCost(cost, n, W));
    }
}
// This code is contributed by Ryuga

PHP

j means capacity of bag
            // is less then weight of item
            if ($wt[$i - 1] > $j)
                $min_cost[$i][$j] = $min_cost[$i - 1][$j];
 
            // here we check we get minimum
            // cost either by including it
            // or excluding it
            else
                $min_cost[$i][$j] = min($min_cost[$i - 1][$j],
                                        $min_cost[$i][$j - $wt[$i - 1]] +
                                                           $val[$i - 1]);
        }
    }
 
    // exactly weight W can not be made
    // by given weights
    if ($min_cost[$n][$W] == $INF)
            return -1;
    else
        return $min_cost[$n][$W];
}
 
// Driver Code
$cost = array(1, 2, 3, 4, 5);
$W = 5;
$n = sizeof($cost);
echo MinimumCost($cost, $n, $W);
 
// This code is contributed by ita_c
?>

Javascript

    // Javascript program to find minimum cost to get exactly
    // W Kg with given packets
     
    let INF = 1000000;
     
    // cost[] initial cost array including unavailable packet
    // W capacity of bag
    function MinimumCost(cost, n, W)
    {
        // val[] and wt[] arrays
        // val[] array to store cost of 'i' kg packet of orange
        // wt[] array weight of packet of orange
        let val = [], wt = [];
 
        // traverse the original cost[] array and skip
        // unavailable packets and make val[] and wt[]
        // array. size variable tells the available number
        // of distinct weighted packets
        let size = 0;
        for (let i=0; i<n; i++)
        {
            if (cost[i]!= -1)
            {
                val.push(cost[i]);
                wt.push(i+1);
                size++;
            }
        }
 
        n = size;
        let min_cost = new Array(n+1);
        for(let i = 0; i < n + 1; i++)
        {
            min_cost[i] = new Array(W + 1);
        }
 
        // fill 0th row with infinity
        for (let i=0; i<=W; i++)
            min_cost[0][i] = INF;
 
        // fill 0'th column with 0
        for (let i=1; i<=n; i++)
            min_cost[i][0] = 0;
 
        // now check for each weight one by one and fill the
        // matrix according to the condition
        for (let i=1; i<=n; i++)
        {
            for (let j=1; j<=W; j++)
            {
                // wt[i-1]>j means capacity of bag is
                // less then weight of item
                if (wt[i-1] > j)
                    min_cost[i][j] = min_cost[i-1][j];
 
                // here we check we get minimum cost either
                // by including it or excluding it
                else
                    min_cost[i][j] = Math.min(min_cost[i-1][j],
                         min_cost[i][j-wt[i-1]] + val[i-1]);
            }
        }
 
        // exactly weight W can not be made by given weights
        return (min_cost[n][W]==INF)? -1: min_cost[n][W];
    }
     
    // Driver code
    let cost = [1, 2, 3, 4, 5], W = 5;
    let n = cost.length;
  
    document.write(MinimumCost(cost, n, W));
     
    // This code is contributed by suresh07.

---

输出

5

空间优化解决方案如果我们仔细研究这个问题，我们可能会注意到这是杆切割问题的一种变体。除了进行最大化之外，我们还需要进行最小化。

C++

// C++ program to find minimum cost to
// get exactly W Kg with given packets
#include
using namespace std;
 
/* Returns the best obtainable price for
   a rod of length n and price[] as prices
   of different pieces */
int minCost(int cost[], int n)
{
   int dp[n+1];
   dp[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 min_cost = INT_MAX;
       for (int j = 0; j < i; j++)
         if(j < n)
             min_cost = min(min_cost, cost[j] + dp[i-j-1]);
       dp[i] = min_cost;
   }
   
   return dp[n];
}
 
/* Driver code */
int main()
{
   int cost[] = {20, 10, 4, 50, 100};
   int W = sizeof(cost)/sizeof(cost[0]);
   cout << minCost(cost, W);
   return 0;
}

Java

// Java program to find minimum cost to
// get exactly W Kg with given packets
import java.util.*;
class Main
{
   
    /* Returns the best obtainable price for
    a rod of length n and price[] as prices
    of different pieces */
    public static int minCost(int cost[], int n)
    {
       int dp[] = new int[n + 1];
       dp[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 min_cost = Integer.MAX_VALUE;
           for (int j = 0; j < i; j++)
               if(j < cost.length) {
                 min_cost = Math.min(min_cost, cost[j] + dp[i - j - 1]);
            }
            dp[i] = min_cost;
       }
        
       return dp[n];
    }
 
    public static void main(String[] args) {
       int cost[] = {20, 10, 4, 50, 100};
       int W = cost.length;
       System.out.print(minCost(cost, W));
    }
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to find minimum cost to
# get exactly W Kg with given packets
import sys
 
# Returns the best obtainable price for
# a rod of length n and price[] as prices
# of different pieces
def minCost(cost, n):
     
    dp = [0 for i in range(n + 1)]
 
    # Build the table val[] in bottom up
       # manner and return the last entry
       # from the table
    for i in range(1, n + 1):
        min_cost = sys.maxsize
 
        for j in range(i):
            if j

C＃

// C# program to find minimum cost to
// get exactly W Kg with given packets
using System;
class GFG {
     
    /* Returns the best obtainable price for
    a rod of length n and price[] as prices
    of different pieces */
    static int minCost(int[] cost, int n)
    {
       int[] dp = new int[n + 1];
       dp[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 min_cost = Int32.MaxValue;
           for (int j = 0; j < i; j++)
             if(j < n)
                 min_cost = Math.Min(min_cost,
                                    cost[j] + dp[i - j - 1]);
           dp[i] = min_cost;
       }
         
       return dp[n];
    }
    
  // Driver code
  static void Main() {
   int[] cost = {20, 10, 4, 50, 100};
   int W = cost.Length;
   Console.Write(minCost(cost, W));
  }
}
 
// This code is contributed by divyesh072019

Java脚本

 
    // Javascript program to find minimum cost to
    // get exactly W Kg with given packets
     
    /* Returns the best obtainable price for
     a rod of length n and price[] as prices
     of different pieces */
    function minCost(cost, n)
    {
       let dp = new Array(n+1);
       dp[0] = 0;
 
       // Build the table val[] in bottom up
       // manner and return the last entry
       // from the table
       for (let i = 1; i<=n; i++)
       {
           let min_cost = Number.MAX_VALUE;
           for (let j = 0; j < i; j++)
             if(j < n)
                 min_cost = Math.min(min_cost, cost[j] + dp[i-j-1]);
           dp[i] = min_cost;
       }
 
       return dp[n];
    }
 
    let cost = [20, 10, 4, 50, 100];
    let W = cost.length;
    document.write(minCost(cost, W));
     

输出

14