HDU 1075 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): 25869 Accepted Submission(s): 8813

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.

Author

Ignatius.L

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题目大意：将给出的火星字符串翻译成英文字符串，每个单词有对应的英文单词，如果不存在对应的就输出原有的即可。

解题思路：对火星字符串进行建树，在最后一个字母上存下对应的英文字符串。这里要注意一个标记，不然会超内存。

如图。画了两个具体说明一下，其余的以此类推。

  1 #include <iostream>
  2 #include <cstring>
  3 #include <string>
  4 #include <map>
  5 #include <algorithm>
  6 #include <cctype>
  7 using namespace std;
  8 typedef struct nn
  9 {
 10     int f;
 11     char *c;
 12     struct nn *nxt[26];
 13 }node;
 14 
 15 node *builde()
 16 {
 17     node*p = (node *)malloc(sizeof(node));
 18     p->f = 0;
 19     for (int i = 0; i < 26; i++)
 20     {
 21         p->nxt[i] = NULL;
 22     }
 23     return p;
 24 }
 25 
 26 node *root,*p;
 27 
 28 void insert(char*t, char *s)//s是原语言，t是英文
 29 {
 30     p = root;
 31     int i;
 32     for (i = 0; s[i] != ''; i++)
 33     {
 34         if (p->nxt[s[i] - 'a'] == NULL)
 35         {
 36             p->nxt[s[i] - 'a'] = builde();
 37         }
 38         p = p->nxt[s[i] - 'a'];
 39     }
 40     p->f = 1;
 41     int l = strlen(t);
 42     p->c = (char *)malloc((l+1)*sizeof(char)); // 申请动态空间
 43     strcpy(p->c,t);
 44 }
 45 
 46 void search(char *s)
 47 {
 48     p = root;
 49     if (strlen(s) == 0) return;
 50     for (int i = 0; s[i] != '';i++)
 51     {
 52         if (p->nxt[s[i] - 'a'] == NULL)
 53         {
 54             cout << s;
 55             return;
 56         }
 57         p = p->nxt[s[i] - 'a'];
 58     }
 59     if (p->f)
 60         cout << p->c;
 61     else
 62         cout << s;
 63 }
 64 
 65 int main()
 66 {
 67     char t[15], s[3005], s1[3005];
 68     root = builde();
 69     p=builde();
 70     while (cin >> s)
 71     {
 72         if (strcmp(s,"START")==0) continue;
 73         if (strcmp(s, "END") == 0)
 74             break;
 75         cin >> s1;
 76         insert(s, s1);
 77     }
 78     scanf("%s", &t);
 79     getchar();
 80     while (gets(s))
 81     {
 82         if (strcmp(s, "END") == 0) break;
 83         int i;
 84         int k=0;
 85         for (i = 0; s[i] != ''; i++)
 86         {
 87             if (s[i]<'a' || s[i]>'z')
 88             {
 89                 s1[k] = '';
 90                 search(s1);
 91                 cout << s[i];
 92                 k = 0;
 93             }
 94             else
 95                 s1[k++] = s[i];
 96         }
 97         cout << endl;
 98     }
 99     return 0;
100 }

用map更方便

 1 #include <iostream>
 2 #include <cstring>
 3 #include <string>
 4 #include <map>
 5 #include <algorithm>
 6 #include <cctype>
 7 using namespace std;
 8 int main()
 9 {
10     map<string,string>m;
11     string a;
12     string s;
13     cin >> s;
14     while (cin >> a && a != "END")
15     {
16         cin >> s;
17         m[s] = a;
18     }
19     cin >> s;
20     getchar();
21     char c[3005];
22     while (1)
23     {
24         gets(c);
25         int i, l;
26         if (strcmp(c, "END") == 0) break;
27         l = strlen(c);
28         s = "";
29         for (i = 0; i < l; i++)
30         {
31             if (c[i]<'a' || c[i]>'z')
32             {
33                 if (m[s] != "") cout << m[s];
34                 else cout << s;
35                 s = "";
36                 cout << c[i];
37             }
38             else s = s + c[i];
39         }
40         cout << endl;
41     }
42     return 0;
43 }