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# -*- coding: utf-8 -*-
import time
def cal_time(func):
def wrapper(*args, **kwargs):
t1 = time.time()
result = func(*args, **kwargs)
t2 = time.time()
print("%s running time: %s secs." % (func.__name__, t2 - t1), end='--------------------')
return result
return wrapper
p = [0, 1, 5, 8, 9, 10, 17, 17, 20, 21, 23, 24, 26, 27, 27, 28, 30, 33, 36, 39, 40]
# p = [0, 1, 5, 8, 9, 10, 17, 17, 20, 24, 30]
# 递归--计算效率低
def cut_rod_recurision_1(p, n):
if n == 0:
return 0
else:
res = p[n]
for i in range(1, n):
res = max(res, cut_rod_recurision_1(p, i) + cut_rod_recurision_1(p, n - i))
return res
@cal_time
def c1(p, n):
return cut_rod_recurision_1(p, n)
def cut_rod_recurision_2(p, n):
if n == 0:
return 0
else:
res = 0
for i in range(1, n + 1):
res = max(res, p[i] + cut_rod_recurision_2(p, n - i))
return res
@cal_time
def c2(p, n):
return cut_rod_recurision_2(p, n)
print(c1(p, 15))
print(c2(p, 15))
# 自底向上实现
@cal_time
def cut_rod_dp(p, n):
r = [0]
for i in range(1, n + 1):
res = 0
for j in range(1, i + 1):
res = max(res, p[j] + r[i - j])
r.append(res)
return r[n]
print(cut_rod_dp(p, 15))
# 重构解
def cut_rod_extend(p, n):
r = [0]
s = [0]
for i in range(1, n + 1):
res_r = 0 # 价格的最大值
res_s = 0 # 价格最大值对应方案的左边不切割部分的长度
for j in range(1, i + 1):
if p[j] + r[i - j] > res_r:
res_r = p[j] + r[i - j]
res_s = j
r.append(res_r)
s.append(res_s)
return r[n], s
print("=================================重构解=================================")
r, s = cut_rod_extend(p, 20)
print(s)
# 切割方案
def cut_rod_solution(p, n):
r, s = cut_rod_extend(p, n)
ans = []
while n > 0:
ans.append(s[n])
n -= s[n]
return ans
print("================================按如下切割===============================")
print(cut_rod_solution(p, 20))