The basic problem of dynamic programming (python achieve)

First, the knapsack problem

1. Description of the problem

2, the introduction of dynamic programming table to explain the problem

This table

Backpack capacity: maximum input is given,

Item number: 0, 0 indicates the best combination of the first article;

　　　　　1, 1 indicates the best combination of the first item; (so there is not a certain number of items requested order ??)

　　　　　2, represents the best combination of the first two items;

The best combination of value can be derived: Blank grid

3, dynamic programming to fill out the form value data

　　a, item number is 0, the first 0 items combined (no items), no matter how much capacity backpack, valued at zero.

　　B, knapsack capacity is 0, the article can not pretend to be, the value of 0

　　C, a capacity of a knapsack, Item No. 1:　　

　　　　　　　　　　　　　　　　(1) No. 1 if fitted items, the value of change;

　　　　　　　　　　　　　　　　(2) If you do not install the 1st item, the value 0 and the value of goods before think the same;

　　　　　　　　　　　　　　　　Since the fit article 1, only select all (2)

　　d, backpack capacity of 2, items numbered 1:

　　　　(1) determine the current capacity of the backpack can fit in the current article; can

　　　　(2) if: the judge hold that the total value of goods and do not install the item total value of the larger

　　　　　　 If you can not: do not install this item, the total value and the total value is equal to the previous line

　　d, backpack capacity of 3, item number 2 is: (1) determine the current capacity of the backpack can fit in the current article; can

　　　　　　　　　　　　　　　　 (2) if: the judging means after the total value of the article (4) is not installed and the total value of the item (3) which is larger

　　　　　　　　　　　　　　　　　　　If you can not: do not install this item, the total value and the total value is equal to the previous line

　　　　　　　　　　　　　　　　　　 Finally, taking the total value of 4 large

　　e, other blank and so on, not detailed. . . .

Code:

1, the data input:

n = int(input())
w = list(map(int,input().split()))
v = list(map(int,input().split()))
c = int(input())

"""
测试数据：
n = 4  物品的数量，
w = [2, 3, 4, 5] 每个物品的重量，
v = [3，4, 5, 6 ] 每个物品的价值，
c = 8 书包能承受的重量，
"""

2、动态规划表的绘制

value = [[0 for j in range(c + 1)] for i in range(n + 1)]  #构造一个dp表
for i in range(1, n + 1):
    for j in range(1, c + 1):
        if (w[i-1]>j):                                     #如果当前物品质量大于当前背包容量  
            value[i][j] = value[i-1][j]                    #不装改物品，总价值和前一个总价值相等
        else:                                              #如果可以装下该物品，比较 不装该物品的价值 和 装下该物品后的价值那个更大，取大值。
            value[i][j] = max(value[i-1][j],value[i-1][j-w[i-1]]+v[i-1])
for x in value:
    print(x)

4、背包问题回溯

理解：根据动态规划表的构建，如果当前物品被放入背包中，则当前总价值大于前一行的总价值。

　　　所以可以根据两行总价值来判断当前物品是否在背包中。

代码实现：

print('最大价值为:', value[n][c])
x = [False for i in range(n)]
j = c
for i in range(n, 0, -1):
    if value[i][j] > value[i - 1][j]:  #判断当前价值与前一行的价值
        x[i - 1] = True                #当前物品置一
        j -= w[i - 1]                  #背包容量里面减去相应的物品容量
print('背包中所装物品为:')
for i in range(n):
    if x[i]:
        print('第', i+1, '个,', end='')

The basic problem of dynamic programming (python achieve)

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