给你一袋 W kg 大小的袋子，并在数组 cost[] 中为你提供不同重量的橙子包的成本，其中cost[i]基本上是“i”公斤橙子包的成本。其中 cost[i] = -1 表示“i”公斤橙包不可用
找出购买恰好 W kg 橙子的最低总成本，如果不可能购买恰好 W kg 的橙子，则打印 -1。可以假设所有可用数据包类型的供应是无限的。
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.这个问题可以简化为Unbounded Knapsack。所以在成本数组中，我们首先忽略那些不可用的数据包，即；成本为-1，然后遍历成本数组并创建两个数组val[]用于存储‘ i’kg橙色包的成本和wt[]用于存储相应包的重量。假设 cost[i] = 50，所以数据包的权重为 i，成本为 50。
算法 ：
- 创建矩阵 min_cost[n+1][W+1]，其中 n 是不同的橙色加权包的数量，W 是包的最大容量。
- 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 那么 min_cost[i][j] = min_cost[i-1][j] ;
- 如果 wt[i-1] <= j 那么 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，因为这意味着我们不能使用这些权重使权重 W 否则输出将是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
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 && cost[j]!=-1)
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 && cost[j]!=-1) {
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[] = {10,-1,-1,-1,-1};
int W = cost.length;
System.out.print(minCost(cost, W));
}
}
// This code is contributed by divyeshrabadiya07Python3
# 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 jC#
// 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 && cost[j]!=-1)
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 divyesh072019Javascript
输出
5空间优化解决方案如果我们仔细观察这个问题，我们可能会注意到这是杆切割问题的一个变体。这里我们需要做最小化而不是最大化。
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 && cost[j]!=-1)
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 && cost[j]!=-1) {
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[] = {10,-1,-1,-1,-1};
int W = cost.length;
System.out.print(minCost(cost, W));
}
}
// This code is contributed by divyeshrabadiya07
蟒蛇3
# 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 jC＃
// 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 && cost[j]!=-1)
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
Javascript
输出
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