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  • poj 2503 Babelfish(字典树哈希)

    Time Limit: 3000MS Memory Limit: 65536K
    Total Submissions: 29059 Accepted: 12565

    Description

    You have just moved from Waterloo to a big city. The people here speak an incomprehensible dialect of a foreign language. Fortunately, you have a dictionary to help you understand them.

    Input

    Input consists of up to 100,000 dictionary entries, followed by a blank line, followed by a message of up to 100,000 words. Each dictionary entry is a line containing an English word, followed by a space and a foreign language word. No foreign word appears more than once in the dictionary. The message is a sequence of words in the foreign language, one word on each line. Each word in the input is a sequence of at most 10 lowercase letters.

    Output

    Output is the message translated to English, one word per line. Foreign words not in the dictionary should be translated as "eh".

    Sample Input

    dog ogday
    cat atcay
    pig igpay
    froot ootfray
    loops oopslay
    
    atcay
    ittenkay
    oopslay
    

    Sample Output

    cat
    eh
    loops

    题意:给一个字典,输入若干个字符串,问再字典中与它对应的字符串,若没有输出“eh”;

     1 #include<stdio.h>
     2 #include<string.h>
     3 #include<stdlib.h>
     4 char s1[100010][20],s2[100010][20];
     5 
     6 struct node
     7 {
     8     int flag;
     9     struct node *next[26];
    10 };
    11 struct node* creat()
    12 {
    13     struct node *p = (struct node *)malloc(sizeof(struct node));
    14     p->flag = 0;
    15     for(int i = 0; i < 26; i++)
    16         p->next[i] = NULL;
    17     return p;
    18 }
    19 
    20 void insert(struct node *p, char s[],int cnt)
    21 {
    22     for(int i = 0; s[i]; i++)
    23     {
    24         if(p->next[s[i]-'a'] == NULL)
    25             p->next[s[i]-'a'] = creat();
    26         p = p->next[s[i]-'a'];
    27     }
    28     p->flag = cnt;
    29 }
    30 int search(struct node *p, char s[])
    31 {
    32     for(int i = 0; s[i]; i++)
    33     {
    34         if(p->next[s[i]-'a'] == NULL)
    35            return -1;
    36         p = p->next[s[i]-'a'];
    37     }
    38     return p->flag;
    39 }
    40 int main()
    41 {
    42     char s[20],str[20];
    43     int cnt = 0;
    44     struct node *root;
    45     root = creat();
    46     while(gets(s))
    47     {
    48         if(strcmp(s,"") == 0)
    49             break;
    50         sscanf(s,"%s %s",s1[cnt],s2[cnt]);
    51         insert(root,s2[cnt],cnt);
    52         cnt++;
    53     }
    54     while(scanf("%s",str)!= EOF)
    55     {
    56         int f = search(root,str);
    57         if(f == -1)
    58             printf("eh
    ");
    59         else printf("%s
    ",s1[f]);
    60     }
    61     return 0;
    62 }
    View Code



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  • 原文地址:https://www.cnblogs.com/LK1994/p/3378161.html
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