POJ2503 哈希（字符串编码）

Babelfish
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 37822 Accepted: 16111
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：
给出多个单词组成的字典，有几个查询，问每个查询是否能找到对应的单词。
题解：
这题用map也能水过，不过时间耗时太高。正确的做法是hash。把字符串编码为一个键值放到哈希表里，然后查询。常用的字符串编码都试了一遍，对应耗时标在编码方式上面。

#include <iostream>#include <algorithm>#include <cstdio>#include <map>#include <cstdlib>#include <cstring>#include <string>#define f(i,a,b) for(int i = a;i<=b;i++)#define fi(i,a,b) for(int i = a;i>=b;i--)#define M 100003#define HASH APhashusing namespace std;struct node{    int hash;    struct node* next;}*link[M] = {NULL};char word[100000][11],dialect[100000][11],qu[11];unsigned int APhash(char* str){    unsigned int hash = 0;    int i;    for(i = 0;*str;i++)        if((i&1)==0)            hash^=((hash<<7)^(*str++)^(hash>>3));        else            hash^=(~((hash<<11)^(*str)^(hash>>5)));    return (hash&0x7FFFFFFF)%M;}//  TLEunsigned int DJBhash(char* str){    unsigned int hash = 5381;    while(*str){        hash += (hash<<5) + (*str);    }    return (hash&0x7FFFFFFF)%M;}//  688msunsigned int PJWhash(char* str){    unsigned int BitsInUnignedInt = (unsigned int)(sizeof(unsigned int)*8);    unsigned int ThreeQuarters = (unsigned int)((BitsInUnignedInt*3)/4);    unsigned int OneEighth = (unsigned int)(BitsInUnignedInt/8);    unsigned int HighBits = (unsigned int)(0xffffffff)<<(BitsInUnignedInt - OneEighth);    unsigned int hash = 0;    unsigned int test = 0;    while(*str){        hash = (hash<<OneEighth)+(*str++);        if((test = hash&HighBits)!=0)            hash = ((hash^=(test>>ThreeQuarters))&(~HighBits));    }    return (hash&0x7FFFFFFF)%M;}//  688msunsigned int JShash(char* str){    unsigned int hash = 1315423911;    while(*str){        hash^=((hash<<5)+(*str++)+(hash>>2));    }    return (hash&0x7FFFFFFF)%M;}//  704msunsigned int RShash(char* str){    unsigned int b = 378551;    unsigned int a = 63689;    unsigned int hash = 0;    while(*str){        hash = hash*a + (*str++);        a*=b;    }    return (hash&0x7FFFFFFF)%M;}//  719msunsigned int SDBMhash(char* str){    unsigned int hash = 0;    while(*str)        hash = (*str++) + (hash<<6) + (hash<<16) - hash;    return (hash&0x7FFFFFFF)%M;}//  688msunsigned int BKDRhash(char* str){    unsigned int seed = 1313131; //31 131 1313 13131 131313    unsigned int hash = 0;    while(*str)        hash = hash*seed + (*str++);    return (hash&0x7FFFFFFF)%M;}//   641ms 常用 linux文件int ELFhash(char* key){    unsigned long h = 0,g;    while(*key){        h = (h<<4) + *key++;        g = h&0xf0000000L;        if(g)            h^=g>>24;        h&=~g;    }    return h%M;}int main(){//    freopen("data.in","r",stdin);    int n = 0;    struct node* p;    while(1){        char c;        c = getchar();        if(c == '\n') break;        else word[n][0] = c;        scanf("%s %s",word[n]+1,dialect[n]);        getchar();        int e = HASH(dialect[n]);        p = (struct node*) malloc(sizeof(struct node));        p->hash = n;        p->next = link[e];        link[e] = p;        n++;    }    while(1){        char c;        c = getchar();        if(c == -1) break;        else qu[0] = c;        scanf("%s",qu+1);        getchar();        int e = HASH(qu);        p = link[e];        while(p!=NULL){            if(strcmp(qu,dialect[p->hash])==0)                break;            p = p -> next;        }        if(p == NULL)            printf("eh\n");        else            printf("%s\n",word[p->hash]);    }    return 0;}