01背包
每种物品就有一个get
C = [3,2,6,7,1,4,9,5]#cost 单个物品所占容量
V = [6,3,5,8,3,1,6,9]#每一个物品的价值
target = 15 #背包容量
F = [0 for i in range(0,target+1)] #初始化 元素个数为背包大小加1（target+1）
n = len(C)sed
def ZeroOneBackPack(cost,value):
for i in reversed(range(cost,target+1)): #逆序遍历
F[i] = max(F[i],F[i-cost]+value)遍历
for i in range(0,n):
ZeroOneBackPack(C[i],V[i])
print (F[target])co
彻底背包
每种物品有无数多个let
C = [3,2,6,7,1,4,9,5]
V = [6,3,5,8,3,1,6,9]
target = 15
F = [0 for i in range(0,target+1)]
n = len(C)tar
def CompleteBackPack(cost,value):
for i in range(cost,target+1):#这是和01背包惟一的区别 正序遍历
F[i] = max(F[i],F[i-cost]+value)
for i in range(0,n): CompleteBackPack(C[i],V[i]) print (F[target])