本博客將用Python 3 實現最小劃分求解:給出一個正整數數組,寫一個程序把這個整數數組分成S1跟S2兩部分,使S1中的和跟S2中的和的絕對值最小。如果有一個一個整數數組 S
有 n
個數,如果Subset1有 m
個數,Subset2必須有 n-m
個數並且 abs(sum(Subset1) – sum(Subset2))
應該最小
這是面試中常見的題型之一,體現的正是揹包算法的思想。
#coding=utf-8
'''
Author:ZhangLongQi
Date:2018-4-1
E-mail:[email protected]
Environment:Python 3.6
Descrip:最小劃分,揹包思想
Improve:優化實現過程
'''
def bag(nums,cap,wet=[],val=[]):
odd=[]
def prints(path,i,j):
if i>0 and j>0:
if path[i][j]==1:
odd.append(i)
prints(path,i-1,j-wet[i])
else:
prints(path,i-1,j)
res=[[0 for i in range(cap+1)] for j in range(nums+1)]
path=[[0 for i in range(cap+1)] for j in range(nums+1)]
for i in range(1,nums+1):
for j in range(1,cap+1):
res[i][j]=res[i-1][j]
if wet[i]<=j:
tmp= val[i]+res[i-1][j-wet[i]]
if tmp>res[i-1][j]:
res[i][j]=tmp
path[i][j]=1
#for i in range(len(path)):
#print(path[i])
prints(path,nums,cap)
return odd,res[nums][cap]
def dsp(k=[]):
s=0
for i in k:
s=s+w[i]
return abs((sum(w)-s)-s)
if __name__=='__main__':
w=[2,3,1,4,6,5]
print('w=',w)
c1=sum(w)//2
c2=sum(w)//2+1 #僅爲了比較
n=len(w)
val=[1 for _ in range(n)]
w.insert(0,0)
val.insert(0,0)
k1,v=bag(n,c1,w,val)
k2,v=bag(n,c2,w,val)
k=k1 if dsp(k1)<dsp(k2) else k2 #Python中沒有三目運算符
print('把第',sorted(k),'個數取出,差值爲',dsp(k))
Demon:
原創,轉載請標明