动态规划之最长公共子序列(LCS)

# -*- coding: utf-8 -*-
import time
gk = lambda i,j:str(i)+','+str(j)
def LSC_length(x, y):
m = len(x)
n = len(y)
b, c = {}, {} # b用来存储解决的方案， c用来存储x序列前缀i和y序列前缀j对应的LCS的长度
for i in xrange(0, m):
c[gk(i, -1)] = 0
for j in xrange(0, n):
c[gk(-1, j)] = 0
for i in xrange(0, m):
for j in xrange(0, n):
if x[i] == y[j]:
c[gk(i, j)] = c[gk(i-1, j-1)]+1
b[gk(i, j)] = 'hit'
elif c[gk(i-1, j)] >= c[gk(i, j-1)]:
c[gk(i, j)] = c[gk(i-1, j)]
b[gk(i, j)] = "fromUp"
else:
c[gk(i, j)] = c[gk(i, j-1)]
b[gk(i, j)] = "fromLeft"
return c, b
def print_LCS(b, x, i, j):
'''将b从最后一个元素顺着箭头往前数就会得到LCS的输出'''
if i == -1 or j == -1:
return
if b[gk(i, j)] == 'hit':
print_LCS(b, x, i-1, j-1)
print x[i]
elif b[gk(i, j)] == 'fromUp':
print_LCS(b, x, i-1, j)
else:
print_LCS(b, x, i, j-1)
def test():
x = ['a','b','c','b','d','a','b']
y = ['b','d','c','a','b','a']
c, b = LSC_length(x, y)
print 'length of LSC of x and y is', c[gk(len(x)-1, len(y)-1)]
print_LCS(b, x, len(x)-1, len(y)-1)

if __name__ == '__main__':
begin = time.time()
test()
print 'total runtime is', time.time()-begin

>>>
length of LSC of x and y is 4
b
c
b
a
total runtime is 0.0339999198914