2017微软秋季校园招聘在线编程笔试-#1400 : Composition

http://hihocoder.com/problemset/problem/1400

题意理解：有限制的两个不能相邻，i之和前面剩下的那一个相关，所以可以f【i】【ch】的动态规划去做，每次转移考虑删不删i的字符

急转弯：想到i只和前面一个相关，并且可以记录下来。

算法：无

数据结构：无

#
#

'f'

__author__ = 'hjkruclion'

import sys

def read_int():
return list(map(int, sys.stdin.readline().split()))
def read_str():
return sys.stdin.readline().split()[0]

maxn = 100000 + 100
MAXNUM = 27
INF = int(1e8)

f = [[INF for j in range(MAXNUM)] for i in range(maxn)]
h = [[0 for j in range(MAXNUM)] for i in range(MAXNUM)]

def no(ch):
return ord(ch) - ord('a')
# print(no('a'))
N = read_int()[0]
t = read_str()
s = ['a' for _ in range(maxn)]
for i in range(1, N + 1):
s[i] = t[i - 1]
M = read_int()[0]
for i in range(1, M + 1):
t = read_str()
h[no(t[0])][no(t[1])] = 1
h[no(t[1])][no(t[0])] = 1

f[0][26] = 0
for i in range(1, N + 1):
t = no(s[i])
for j in range(0, 27):
f[i][j] = min(f[i][j], f[i - 1][j] + 1)
for j in range(0, 27):
if h[t][j] == 0:
f[i][t] = min(f[i][t], f[i - 1][j])
res = INF
for i in range(0, 27):
res = min(res, f
[i])
print(res)