POJ2503-Babelfish (三种解法)

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;
}