2016微软探星夏令营在线技术笔试-#1341 : Constraint Checker

来源：互联网发布：算法公平和效率知乎编辑：程序博客网时间：2024/05/18 02:46

http://hihocoder.com/problemset/problem/1341

题意理解：按照描述的每次转换字母然后判定就好了

急转弯：有点麻烦

算法：无

数据结构：无

from __future__ import print_function##'interpret'__author__ = 'hjkruclion'maxn = 30maxnLen = 1e5a = [[] for i in range(maxn)]flag = [[] for i in range(maxn)]cTOn = [0 for i in range(1000)]has = [0 for i in range(1000)]import sysdef read_int():    return list(map(int, sys.stdin.readline().split()))def read_str():    return sys.stdin.readline().split()[0]def read_white_str():    return sys.stdin.readline().split()def isdigit(ch):    if ord(ch) >= ord('0') and ord(ch) <= ord('9'):        return True    return Falsedef changeA(t):    if isinstance(t, int):        return t    return cTOn[ord(t)]cnt = 0N = read_int()[0]for i in range(N):    t = read_str()    l = len(t)    j = 0    while(j < l):        if t[j] == '<' and t[j + 1] != '=':            flag[i].append(1)            j += 1        elif t[j] == '<' and t[j + 1] == '=':            flag[i].append(2)            j += 2        elif isdigit(t[j]):            k = j            while(k < l and isdigit(t[k])):                k += 1            num = int(t[j:k])            a[i].append(num)            j = k        else:            if has[ord(t[j])] == 0:                has[ord(t[j])] = 1                cnt += 1            a[i].append(t[j])            j += 1T = read_int()[0]for _ in range(T):    for i in range(cnt):        t = read_white_str()        cTOn[ord(t[0][0])] = int(t[1])    ans = 1    for i in range(N):        l = len(a[i])        for j in range(1, l):            x = changeA(a[i][j - 1])            y = changeA(a[i][j])            if flag[i][j - 1] == 1:                if y <= x:                    ans = 0                    break            if flag[i][j - 1] == 2:                if y < x:                    ans = 0                    break    if ans == 1:        print('Yes')    else:        print('No')

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