给定一个大小为n的数组包含糖果的成本，以使得sweet [i]是i糖果的成本。任务是找到为“ m”个人购买恰好“ n”公斤甜食的最低花费。
由于小包中有糖果，因此您必须为“ m”个亲戚购买最多“ m”个小包。您不能购买超过“ m”包的糖果。同样cost [i] = 0，表示具有数据包大小i的sweet不可用。同样，有无数个大小为i的小包。
例子：
Input: m = 3, n = 6, arr[] = {2, 1, 3, 0, 4, 10}
Output: 3
We can choose at most 3 packets. We choose 3 packets of size 2 having cost 1 each.Thus output is 3.
Input: m = 2, n = 7, arr[] = {1, 3, 0, 5, 0, 0, 0}
Output : 0
We can choose at most 2 packets. 7 can be formed by 1 2 and 4 index but since you require at most 2 packets to obtain the 7 sweets packets answer which is not possible. Hence the answer is 0 as it is formed by 3 packets, not 2.
方法：
- 创建矩阵sweet [m + 1] [n + 1] [n + 1]，其中m是亲戚数量，n是要购买的总甜品公斤和甜品包装数量。
- 用0初始化sweet [i] [0] [j]元素，用-1初始化sweet [i] [j] [0]。
- 现在根据以下规则填充矩阵–
- 购买“ k”软件包并将其分配给dp阵列。如果i> 0且j> =当前包装数且k个糖果的价格大于0。将dp定义为d p [i-1] [jk] [k] + sweet [k]
- 如果dp未定义，请从以前的k-1个软件包中选择-> dp [i] [j] [k] = dp [i] [j] [k-1]
- 如果dp [m] [n] [n]为-1，则答案为0，否则打印dp [m] [n] [n]
下面是上述方法的实现：
C++
// C++ program to minimum cost to buy
// N kilograms of sweet for M persons
#include
using namespace std;
// Function to find the minimum cost of sweets
int find(int m, int n, int adj[])
{
// Defining the sweet array
int sweet[n + 1];
// DP array to store the values
int dp[n + 1][n + 1][n + 1];
sweet[0] = 0;
// Since index starts from 1 we
// reassign the array into sweet
for (int i = 1; i <= m; ++i)
sweet[i] = adj[i - 1];
// Assigning base cases for dp array
for (int i = 0; i <= m; ++i) {
for (int k = 0; k <= n; ++k)
// At 0 it is free
dp[i][0][k] = 0;
// Package not available for desirable amount of sweets
for (int k = 1; k <= n; ++k)
dp[i][k][0] = -1;
}
for (int i = 0; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
for (int k = 1; k <= n; ++k) {
dp[i][j][k] = -1;
// Buying the 'k' kg package and
// assigning it to dp array
if (i > 0 && j >= k && sweet[k] > 0
&& dp[i - 1][j - k][k] != -1)
dp[i][j][k] = dp[i - 1][j - k][k] + sweet[k];
// If no solution, select from previous k-1 packages
if (dp[i][j][k] == -1 || (dp[i][j][k - 1] != -1
&& dp[i][j][k] > dp[i][j][k - 1]))
dp[i][j][k] = dp[i][j][k - 1];
}
}
}
// If solution does not exist
if (dp[m][n][n] == -1)
return 0;
// Print the solution
else
return dp[m][n][n];
}
// Driver Function
int main()
{
int m = 3;
int adj[] = { 2, 1, 3, 0, 4, 10 };
int n = sizeof(adj) / sizeof(adj[0]);
// Calling the desired function
cout << find(m, n, adj);
return 0;
} Java
// Java program to minimum cost to buy
// N kilograms of sweet for M persons
public class GFG {
// Function to find the minimum cost of sweets
static int find(int m, int n, int adj[])
{
// Defining the sweet array
int sweet[] = new int [n + 1] ;
// DP array to store the values
int dp[][][] = new int [n + 1][n + 1][n + 1] ;
sweet[0] = 0;
// Since index starts from 1 we
// reassign the array into sweet
for (int i = 1; i <= m; ++i)
sweet[i] = adj[i - 1];
// Assigning base cases for dp array
for (int i = 0; i <= m; ++i) {
for (int k = 0; k <= n; ++k)
// At 0 it is free
dp[i][0][k] = 0;
// Package not available for desirable amount of sweets
for (int k = 1; k <= n; ++k)
dp[i][k][0] = -1;
}
for (int i = 0; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
for (int k = 1; k <= n; ++k) {
dp[i][j][k] = -1;
// Buying the 'k' kg package and
// assigning it to dp array
if (i > 0 && j >= k && sweet[k] > 0
&& dp[i - 1][j - k][k] != -1)
dp[i][j][k] = dp[i - 1][j - k][k] + sweet[k];
// If no solution, select from previous k-1 packages
if (dp[i][j][k] == -1 || (dp[i][j][k - 1] != -1
&& dp[i][j][k] > dp[i][j][k - 1]))
dp[i][j][k] = dp[i][j][k - 1];
}
}
}
// If solution does not exist
if (dp[m][n][n] == -1)
return 0;
// Print the solution
else
return dp[m][n][n];
}
// Driver code
public static void main(String args[])
{
int m = 3;
int adj[] = { 2, 1, 3, 0, 4, 10 };
int n = adj.length ;
System.out.println( find(m, n, adj));
}
// This Code is contributed by ANKITRAI1
}Python
# Python3 program to minimum cost to buy
# N kilograms of sweet for M persons
# Function to find the minimum cost of sweets
def find(m, n, adj):
# Defining the sweet array
sweet = [0] * (n + 1)
# DP array to store the values
dp = [[[ 0 for i in range(n + 1)] for i in range(n + 1)] for i in range(n + 1)]
sweet[0] = 0
# Since index starts from 1 we
# reassign the array into sweet
for i in range(1, m + 1):
sweet[i] = adj[i - 1]
# Assigning base cases for dp array
for i in range(m + 1):
for k in range(n + 1):
# At 0 it is free
dp[i][0][k] = 0
# Package not available for desirable amount of sweets
for k in range(1, n + 1):
dp[i][k][0] = -1
for i in range(m + 1):
for j in range(1, n + 1):
for k in range(1, n + 1):
dp[i][j][k] = -1
# Buying the 'k' kg package and
# assigning it to dp array
if (i > 0 and j >= k and sweet[k] > 0 and dp[i - 1][j - k][k] != -1):
dp[i][j][k] = dp[i - 1][j - k][k] + sweet[k]
# If no solution, select from previous k-1 packages
if (dp[i][j][k] == -1 or (dp[i][j][k - 1] != -1 and dp[i][j][k] > dp[i][j][k - 1])):
dp[i][j][k] = dp[i][j][k - 1]
# If solution does not exist
if (dp[m][n][n] == -1):
return 0
# Print the solution
else:
return dp[m][n][n]
# Driver Function
m = 3
adj = [2, 1, 3, 0, 4, 10]
n = len(adj)
# Calling the desired function
print(find(m, n, adj))
# This code is contributed by mohit kumar 29C#
// C# program to minimum cost to buy
// N kilograms of sweet for M persons
using System;
class GFG
{
// Function to find the minimum
// cost of sweets
static int find(int m, int n,
int[] adj)
{
// Defining the sweet array
int[] sweet = new int [n + 1] ;
// DP array to store the values
int[,,] dp = new int [n + 1, n + 1,
n + 1];
sweet[0] = 0;
// Since index starts from 1 we
// reassign the array into sweet
for (int i = 1; i <= m; ++i)
sweet[i] = adj[i - 1];
// Assigning base cases
// for dp array
for (int i = 0; i <= m; ++i)
{
for (int k = 0; k <= n; ++k)
// At 0 it is free
dp[i, 0, k] = 0;
// Package not available for
// desirable amount of sweets
for (int k = 1; k <= n; ++k)
dp[i, k, 0] = -1;
}
for (int i = 0; i <= m; ++i)
{
for (int j = 1; j <= n; ++j)
{
for (int k = 1; k <= n; ++k)
{
dp[i, j, k] = -1;
// Buying the 'k' kg package and
// assigning it to dp array
if (i > 0 && j >= k && sweet[k] > 0 &&
dp[i - 1, j - k, k] != -1)
dp[i, j, k] = dp[i - 1, j - k, k] +
sweet[k];
// If no solution, select from
// previous k-1 packages
if (dp[i, j, k] == -1 ||
(dp[i, j, k - 1] != -1 &&
dp[i, j, k] > dp[i, j, k - 1]))
dp[i, j, k] = dp[i, j, k - 1];
}
}
}
// If solution does not exist
if (dp[m, n, n] == -1)
return 0;
// Print the solution
else
return dp[m, n, n];
}
// Driver code
public static void Main()
{
int m = 3;
int[] adj = { 2, 1, 3, 0, 4, 10 };
int n = adj.Length ;
Console.Write(find(m, n, adj));
}
}
// This code is contributed
// by ChitraNayal
// C# program to minimum cost to buy
// N kilograms of sweet for M persons
using System;
class GFG
{
// Function to find the minimum
// cost of sweets
static int find(int m, int n,
int[] adj)
{
// Defining the sweet array
int[] sweet = new int [n + 1] ;
// DP array to store the values
int[,,] dp = new int [n + 1, n + 1,
n + 1];
sweet[0] = 0;
// Since index starts from 1 we
// reassign the array into sweet
for (int i = 1; i <= m; ++i)
sweet[i] = adj[i - 1];
// Assigning base cases
// for dp array
for (int i = 0; i <= m; ++i)
{
for (int k = 0; k <= n; ++k)
// At 0 it is free
dp[i, 0, k] = 0;
// Package not available for
// desirable amount of sweets
for (int k = 1; k <= n; ++k)
dp[i, k, 0] = -1;
}
for (int i = 0; i <= m; ++i)
{
for (int j = 1; j <= n; ++j)
{
for (int k = 1; k <= n; ++k)
{
dp[i, j, k] = -1;
// Buying the 'k' kg package and
// assigning it to dp array
if (i > 0 && j >= k && sweet[k] > 0 &&
dp[i - 1, j - k, k] != -1)
dp[i, j, k] = dp[i - 1, j - k, k] +
sweet[k];
// If no solution, select from
// previous k-1 packages
if (dp[i, j, k] == -1 ||
(dp[i, j, k - 1] != -1 &&
dp[i, j, k] > dp[i, j, k - 1]))
dp[i, j, k] = dp[i, j, k - 1];
}
}
}
// If solution does not exist
if (dp[m, n, n] == -1)
return 0;
// Print the solution
else
return dp[m, n, n];
}
// Driver code
public static void Main()
{
int m = 3;
int[] adj = { 2, 1, 3, 0, 4, 10 };
int n = adj.Length ;
Console.Write(find(m, n, adj));
}
}
// This code is contributed
// by ChitraNayal输出：
3
上述算法的时间复杂度为O(m * n * n)。
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