poj 2503 Babelfish

Babelfish

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 40985 Accepted: 17467

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 ogdaycat atcaypig igpayfroot ootfrayloops oopslayatcayittenkayoopslay

Sample Output

catehloops

Hint

Huge input and output,scanf and printf are recommended.

题意是

输入一个字典，字典格式为“英语à外语”的一一映射关系

然后输入若干个外语单词，输出他们的 英语翻译单词，如果字典中不存在这个单词，则输出“eh”

输入输出都是你如果多输出一个回车就会结束的

这题竟然归为了hash所以我用的是hash的方法，其实就用简单的map函数就可以直接用了，map好像是1900MS左右，hash我用的是大约在1600MS左右

代码如下

#include <stdio.h>#include <string.h>#include <algorithm>#include <iostream>#include <vector>#define MAX 14997using namespace std;struct node{    char ch[102];    char ch1[102];};vector<node>v[15000];int hash ( char str1[] ){    int i;    int len = strlen(str1);    int pos = 0;    for ( i = 0; i < len; i++ )    {        pos = pos+(int)str1[i]*(1+ i);    }    return pos%MAX;}int find ( char ch[], char ch1[] ){    if ( strcmp ( ch, ch1) == 0 )    {        return 1;    }    return 0;}int main(){    char str[102], str1[102];    char a[200];    int i;    while ( gets(a) &&a[0]!='\0' )    {        /*Separating element   分离元素*/        sscanf ( a,"%s %s", str, str1) ;        int pos = hash ( str1 );                /*Storage of data   储存数据*/        node temp;        strcpy( temp.ch, str );        strcpy( temp.ch1, str1 );        v[pos].push_back(temp);    }    while ( gets(a) &&a[0]!='\0'  )    {        int pos = hash ( a );                int ok = 1;        for ( i = 0;i < v[pos].size(); i++ )        {                    /*Fine the same elements  查找相同元素*/            if ( find( v[pos][i].ch1, a ) )            {                printf ("%s\n", v[pos][i].ch );                ok = 0;                break;            }        }        if ( ok == 1 )        {            printf("eh\n");        }    }}

用字典树更省时

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;typedef struct DicTrie{    struct DicTrie *next[26];    char word[11];    int isWord;}*Trie;void insertWord( Trie node, char *stc, char *str ){    int id, i;    while ( *str )    {        id = *str-'a';        if ( node->next[id] == NULL )        {            node->next[id] = new DicTrie;            node->next[id]->isWord = 0;            for ( i = 0;i < 26; i++ )            {                node->next[id]->next[i] = NULL;            }        }        node = node->next[id];        str++;    }    node->isWord = 1;    strcpy( node->word, stc );}char* SearchWord(Trie node,char* st){    int id;    while(*st)    {        id=*st-'a';        if(node->next[id]==NULL)        {            return "eh";        }        node=node->next[id];        st++;    }    if(node->isWord)    {        return node->word;    }    else    {        return "eh";    }}int main(){    int i;    char stc[11], str[11], st[25];    Trie node;    node = new DicTrie;    node->isWord = 0;    for ( i = 0;i < 26; i++ )    {        node->next[i] = NULL;    }    while ( gets(st) && st[0] != 0)    {        sscanf ( st,"%s %s", stc, str) ;        insertWord(node, stc, str);    }    while ( ~scanf ( "%s", st ) )    {        char *sum = SearchWord(node, st);        printf ( "%s\n" ,sum );    }}

代码菜鸟，如有错误，请多包涵！！！