POJ2503-Babelfish (三种解法)

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大致题意：

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”。

。

1、直接使用map。。。。 938MS

#include <string>#include <cstdio>#include <cstring>#include <iostream>#include <cstdlib>#include <map>using namespace std;int main(){    char english[11], foreign[11];    map<string, string> translate;    // freopen("in.txt","r",stdin);    while(true) {        char t;        if((t=getchar())=='\n') break;        else {            english[0] = t;            int i = 1;            while(true) {                t = getchar();                if(t == ' ') {                    english[i] = '\0';                    break;                } else                    english[i++] = t;            }        }        scanf("%s",foreign);        getchar();        translate[foreign] = english;    }    char word[11],ans[11];    while(~scanf("%s", word)) {        if(translate.find(word)!=translate.end()) {            strcpy(ans,translate[word].c_str());            printf("%s\n",ans);        } else            printf("eh\n");    }    return 0;}

2 、排序+二分查找 375MS

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;struct word {    char en[11], fl[11];    bool operator < (const word& rhs) const {        return strcmp( fl, rhs.fl)<0;    }};word DICT[100001];int n;int BinarySrearch(char s[], int n){    int l = 0, r = n-1, mid;    while(l<=r) {        mid = (l+r) >> 1;        if( strcmp(s, DICT[mid].fl) == 0) return mid;        else if(strcmp(s, DICT[mid].fl) < 0)            r = mid - 1;        else            l = mid + 1;    }    return -1;}void init(){    char str[30];    n = 0;    while(gets(str)) {        if(str[0]=='\0') break;        sscanf(str,"%s %s", DICT[n].en, DICT[n].fl);        n++;    }    sort(DICT, DICT + n);}void work(){    char temp[11];    int key;    while(gets(temp)) {        key = BinarySrearch(temp, n);        if(-1 == key)            printf("eh\n");        else            printf("%s\n", DICT[key].en);    }}int main(){    init();    work();    return 0;}

3、字符串hash (拉链法) 188MS

#include <stdio.h>#include <string.h>const int M = 200000;typedef struct word{    char e[11];    char f[11];    int next;};word dict[M];int n = 1;int Hash[M];int ELFHash(char* key){    unsigned int  h = 0;    while(*key)    {        h = (h<<4) + (*key++);        unsigned int g = h&(0xf00000000L);        if(g) h^=g>>24;        h&=~g;    }    return h%M;}void find(char f[]){    int h = ELFHash(f);    for(int k=Hash[h]; k; k=dict[k].next)    {        if(strcmp(f, dict[k].f)==0)        {            printf("%s\n",dict[k].e);            return ;        }    }    printf("eh\n");}int main(){    char str[23];    while(gets(str))    {        if(str[0]=='\0') break;        sscanf(str,"%s %s",dict[n].e,dict[n].f);        int h = ELFHash(dict[n].f);        dict[n].next = Hash[h];        Hash[h] = n++;    }    while(gets(str))    {        find(str);    }    return 0;}