给定重量上限,背包问题

Problem statement:

问题陈述：

You are given a bag of size W kg and you are provided costs of packets different weights of oranges in array cost[] where cost[i] is basically cost of i kg packet of oranges. cost[i] = -1 means that i kg packet of orange is unavailable.

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。 cost [i] = -1表示i kg包橙色不可用。

Find the minimum total cost to buy exactly W kg oranges and if it is not possible to buy exactly W kg oranges then print -1. It may be assumed that there is infinite supply of all available packet types.

找到购买正好为W千克橘子的最低总成本，如果不可能购买正好为W千克橘子，则打印-1 。 可以假定存在所有可用分组类型的无限供应。

Input:

输入：

The first line of input contains an integer T denoting the number of test cases. Then T test cases follow. The first line of each test case contains integers N and W where N denotes the size of array cost[ ] and W is the size of the bag.

输入的第一行包含一个整数T，表示测试用例的数量。 然后是T测试用例。 每个测试用例的第一行包含整数N和W ，其中N表示数组cost []的大小， W表示袋子的大小。

The second line of each test case contains N space separated integers denoting elements of the array cost[ ].

每个测试用例的第二行包含N个由空格分隔的整数，它们表示数组cost []的元素。

Output:

输出：

Print the minimum cost to buy exactly W kg oranges. If no such answer exists print "-1".

打印最低成本以购买正好为W公斤的橘子。 如果不存在这样的答案，请打印“ -1” 。

Constraints:

限制条件：

1 <= T <= 100
1 <= N, W <= 1000
1 <= cost[] <= 1000

Example:

例：

Input:
2
5 5
20 10 4 50 100
6 5
-1 -1 4 5 -1 -1
Output:
14
-1

Explanation:

说明：

So, for the first test case,
The 1 kg orange packet costs: 20
The 2 kg orange packet costs: 10
The 3 kg orange packet costs: 4
The 4 kg orange packet costs: 50
The 5 kg orange packet costs: 100
We need total of 5 Kg orange
So, there are many options to pick orange packets
Like picking five 1 kg packets costing 100
two 2 kg and one 1kg packet costing 40
one 2kg and one 3kg packet costing 14
one 1kg and one 4kg costing 70
one 5kg costing 100
so minimum cost one would be one 2kg and
one 3kg bag combination which costs 14
For the second test case,
There is no possible combination to sum total 5kg oranges
Since, only available weight packets are of 3kg and 4kg
We can't take fractional parts
So, it's not possible and answer is -1.

Solution approach

解决方法

This is kind of a duplicate knapsack problem with some constraints. Here we need to keep the check for available packets too.

这是一个重复的背包问题，但有一些约束。 在这里，我们还需要检查可用数据包。

For example,

例如，

Say we have total weights 8

假设我们有总重量8

kg and for an instance, we are checking with packet weight 5

公斤，例如，我们正在检查包装重量5

In such a case, we need to check to things.

在这种情况下，我们需要检查一下事情。

1. Whether the 5kg packet is available or not.

   5公斤小包是否可用。

2. Whether the sub-problem of size (8-5) kg is already solved or not.

   (8-5)kg大小的子问题是否已解决。

If both satisfies then only we can proceed with this instance to compute the current problem.

如果两个都满足，则只有我们可以继续进行本实例计算当前问题。

The above concepts lead to the recurring formula which can be easily converted to Dynamic Programming like below,

上面的概念导致了重复出现的公式，可以很容易地将其转换为动态编程，如下所示，

1. Initialize dp[w+1] with INT_MAX where w is the total weight which is to be computed;

   用INT_MAX初始化dp [w + 1]，其中w是要计算的总权重；

2. Set dp[0]=0 as base case,

   将dp [0] = 0设置为基本情况，

3. for(int i=1;i<=w;i++)
   for(int j=0;j<n;j++)
   if ( a[j]  is availble  && (j+1)<=i && dp[i-(j+1)]  is already computed && dp[i-(j+1)]+a[j]<dp[i])
   dp[i]=dp[i-(j+1)]+a[j];
   end if
   end for
   end for
   
   

4. if(dp[w]  is not computed over the process,contains still the initial value)
   return -1;
   else dp[w] is the minimum cost
   
   

C++ Implementation:

C ++实现：

#include <bits/stdc++.h>
using namespace std;
int minimumCost(vector<int> a, int n, int w)
{
int dp[w + 1];
dp[0] = 0;
for (int i = 1; i <= w; i++)
dp[i] = INT_MAX;
for (int i = 1; i <= w; i++) {
for (int j = 0; j < n; j++) {
if (a[j] != -1 && (j + 1) <= i && dp[i - (j + 1)] != INT_MAX && dp[i - (j + 1)] + a[j] < dp[i]) {
dp[i] = dp[i - (j + 1)] + a[j];
}
}
}
if (dp[w] == INT_MAX)
return -1;
return dp[w];
}
int main()
{
int t, n, item, w;
cout << "Enter number of test cases\n";
cin >> t;
for (int i = 0; i < t; i++) {
cout << "Enter number of elements\n";
cin >> n;
cout << "Enter total weight\n";
cin >> w;
cout << "Enter the cost of weights, -1 if not available\n";
vector<int> a;
for (int j = 0; j < n; j++) {
scanf("%d", &item);
a.push_back(item);
}
cout << "minimum cost is: " << minimumCost(a, n, w) << endl;
}
return 0;
}

Output:

输出：

Enter number of test cases
2
Enter number of elements
5
Enter total weight
5
Enter the cost of weights, -1 if not available
20 10 4 50 100
minimum cost is: 14
Enter number of elements
6
Enter total weight
5
Enter the cost of weights, -1 if not available
-1 -1 4 5 -1 -1
minimum cost is: -1

翻译自: https://www.includehelp.com/icp/minimum-cost-to-fill-given-weight-in-a-bag.aspx

给定重量上限,背包问题