如何解决背包问题经典
| 因此，我必须在上课时解决背包问题。到目前为止，我已经提出了以下建议。我的比较器是确定两个主题中哪个主题更好的函数（通过查看相应的（值，工作）元组）。 我决定对工作量小于maxWork的可能主题进行迭代，为了找出在任何给定回合中哪个主题是最佳选择，我将最近的主题与我们尚未使用的所有其他主题进行了比较。def greedyAdvisor(subjects,maxWork,comparator):
\"\"\"
Returns a dictionary mapping subject name to (value,work) which includes
subjects selected by the algorithm,such that the total work of subjects in
the dictionary is not greater than maxWork. The subjects are chosen using
a greedy algorithm. The subjects dictionary should not be mutated.
subjects: dictionary mapping subject name to (value,work)
maxWork: int >= 0
comparator: function taking two tuples and returning a bool
returns: dictionary mapping subject name to (value,work)
\"\"\"
optimal = {}
while maxWork > 0:
new_subjects = dict((k,v) for k,v in subjects.items() if v[1] < maxWork)
key_list = new_subjects.keys()
for name in new_subjects:
#create a truncated dictionary
new_subjects = dict((name,new_subjects.get(name)) for name in key_list)
key_list.remove(name)
#compare over the entire dictionary
if reduce(comparator,new_subjects.values())==True:
#insert this name into the optimal dictionary
optimal[name] = new_subjects[name]
#update maxWork
maxWork = maxWork - subjects[name][1]
#and restart the while loop with maxWork updated
break
return optimal
解决方法
与OPT相比，使用简单的贪心算法不会对解决方案的质量提供任何限制。
这是一个完整的多项式时间（1-epsilon）*背包的OPT近似伪代码：
items = [...] # items
profit = {...} # this needs to be the profit for each item
sizes = {...} # this needs to be the sizes of each item
epsilon = 0.1 # you can adjust this to be arbitrarily small
P = max(items) # maximum profit of the list of items
K = (epsilon * P) / float(len(items))
for item in items:
profit[item] = math.floor(profit[item] / K)
return _most_prof_set(items,sizes,profit,P)
A(i + 1,p) = min {A(i,p),size(item at i + 1 position) + A(i,p - profit(item at i + 1 position))} if profit(item at i + 1) < p otherwise A(i,p)
def _most_prof_set(items,P):
A = {...}
for i in range(len(items) - 1):
item = items[i+1]
oitem = items[i]
for p in [P * k for k in range(1,i+1)]:
if profit[item] < p:
A[(item,p)] = min([A[(oitem,p)],\\
sizes[item] + A[(item,p - profit[item])]])
else:
A[(item,p)] = A[(oitem,p)] if (oitem,p) in A else sys.maxint
return max(A)
def swap(a,b):
return b,a
def sort_in_decreasing_order_of_profit(ratio,weight,profit):
for i in range(0,len(weight)):
for j in range(i+1,len(weight)) :
if(ratio[i]<ratio[j]):
ratio[i],ratio[j]=swap(ratio[i],ratio[j])
weight[i],weight[j]=swap(weight[i],weight[j])
profit[i],profit[j]=swap(profit[i],profit[j])
return ratio,profit
def knapsack(m,i,newpr):
if(i<len(weight) and m>0):
if(m>weight[i]):
newpr+=profit[i]
else:
newpr+=(m/weight[i])*profit[i]
newpr=knapsack(m-weight[i],i+1,newpr)
return newpr
def printing_in_tabular_form(ratio,profit):
print(\" WEIGHT\\tPROFIT\\t RATIO\")
for i in range(0,len(ratio)):
print (\'{}\\t{} \\t {}\'.format(weight[i],profit[i],ratio[i]))
weight=[10.0,10.0,18.0]
profit=[24.0,15.0,25.0]
ratio=[]
for i in range(0,len(weight)):
ratio.append((profit[i])/weight[i])
#caling function
ratio,profit=sort_in_decreasing_order_of_profit(ratio,profit)
printing_in_tabular_form(ratio,profit)
newpr=0
newpr=knapsack(20.0,newpr)
print(\"Profit earned=\",newpr)
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