hash Babelfish poj 2503

Babelfish

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 27380 Accepted: 11823

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

2503Accepted5068K235MSC++1376B

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define SIZE 199999struct Node {   char key[12], fore[12];   bool flag;   Node () { flag = false; }} table[SIZE];unsigned int BKDRHash(char *str){    unsigned int seed = 131;    unsigned int hash = 0;    while (*str)    {        hash = hash * seed + (*str++);    }    return (hash & 0x7FFFFFFF)%SIZE;}void insert(char *s1, char *s2){    int pos = BKDRHash(s2), m = 0;    while (table[pos].flag){        pos += 2 * (++m) - 1;        if (pos >= SIZE) pos -=SIZE;    }    strcpy(table[pos].key, s2);    strcpy(table[pos].fore, s1);    table[pos].flag = true;}int main(){    char buff[50], s1[20], s2[20];    while (gets(buff) && buff[0] != '\0')    {        sscanf (buff, "%s %s", s1, s2);        insert (s1, s2);    }    while (scanf ("%s", s1) != EOF){        int pos = BKDRHash(s1), m = 0;        bool flag = false;        while (table[pos].flag){            if (strcmp(table[pos].key, s1) == 0) { printf ("%s\n", table[pos].fore); flag = true; break; }            else {                pos += 2 * (++m) - 1;                pos = (pos >=SIZE ? pos -SIZE : pos);            }        }        if (!flag) printf ("eh\n");        getchar();    }    return 0;}