zoukankan      html  css  js  c++  java
  • [HDOJ1075]What Are You Talking About

    What Are You Talking About

    Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/204800 K (Java/Others)
    Total Submission(s): 17546    Accepted Submission(s): 5712


    Problem Description
    Ignatius is so lucky that he met a Martian yesterday. But he didn't know the language the Martians use. The Martian gives him a history book of Mars and a dictionary when it leaves. Now Ignatius want to translate the history book into English. Can you help him?
     
    Input
    The problem has only one test case, the test case consists of two parts, the dictionary part and the book part. The dictionary part starts with a single line contains a string "START", this string should be ignored, then some lines follow, each line contains two strings, the first one is a word in English, the second one is the corresponding word in Martian's language. A line with a single string "END" indicates the end of the directory part, and this string should be ignored. The book part starts with a single line contains a string "START", this string should be ignored, then an article written in Martian's language. You should translate the article into English with the dictionary. If you find the word in the dictionary you should translate it and write the new word into your translation, if you can't find the word in the dictionary you do not have to translate it, and just copy the old word to your translation. Space(' '), tab(' '), enter(' ') and all the punctuation should not be translated. A line with a single string "END" indicates the end of the book part, and that's also the end of the input. All the words are in the lowercase, and each word will contain at most 10 characters, and each line will contain at most 3000 characters.
     
    Output
    In this problem, you have to output the translation of the history book.
     
    Sample Input
    START from fiwo hello difh mars riwosf earth fnnvk like fiiwj END START difh, i'm fiwo riwosf. i fiiwj fnnvk! END
     
    Sample Output
    hello, i'm from mars. i like earth!
    Hint
    Huge input, scanf is recommended.
     
      字典树可以过,不过这个题用STL更明显一些。
      注意cout会超时,printf的话会报一个warnin,如下:
    warning: cannot pass objects of non-POD type `struct std::basic_string<char, std::char_traits<char>, std::allocator<char> >' through `...'; call will abort at runtime

      结果就是导致程序无法正常运行。

      我采用取消cin cout与stdin的同步来解决cout会超时的问题:

        ios::sync_with_stdio(false);

      代码如下:

     1 #include <cstdio>
     2 #include <cstdlib>
     3 #include <cstring>
     4 #include <algorithm>
     5 #include <iostream>
     6 #include <cmath>
     7 #include <queue>
     8 #include <map>
     9 #include <stack>
    10 #include <list>
    11 #include <vector>
    12 #include <cctype>
    13 
    14 using namespace std;
    15 
    16 typedef map<string, string> mss;
    17 typedef pair<string, string> pss;
    18 mss dic;
    19 pss tmp;
    20 char a[12];
    21 char b[12];
    22 char buf[12];
    23 
    24 int main() {
    25     ios::sync_with_stdio(false);
    26     // freopen("in", "r", stdin);
    27     gets(buf);    //START
    28     while(scanf("%s %s", a, b) && strcmp(a, "END")) {    //dic
    29         tmp = make_pair(b, a);
    30         dic.insert(tmp);
    31     }
    32     getchar();
    33     char ch;
    34     int cnt = 0;
    35     while(~scanf("%c", &ch)) {
    36         if(isalpha(ch)) {
    37             buf[cnt++] = ch;
    38         }
    39         else {
    40             buf[cnt] = 0;
    41             cnt = 0;
    42             if(strcmp(buf, "END") == 0) {
    43                 break;
    44             }
    45             if(dic.find(buf) != dic.end()) {
    46                 cout << dic[buf];
    47             }
    48             else {
    49                 cout << buf;
    50             }
    51             cout << ch;
    52         }
    53     }
    54     return 0;
    55 }
  • 相关阅读:
    java多线程之系列目录
    RecyclerView的源码分析
    ConCurrentHashMap在1.7和1.8区别
    插件化之细节
    组件化之开发细节
    组件化之开发总结
    线程之volatile基本内容
    线程之Synchronized基本内容
    设计模式之动态代理模式原理介绍
    操作系统之内存映射
  • 原文地址:https://www.cnblogs.com/kirai/p/4759847.html
Copyright © 2011-2022 走看看