PKU Babelfish

Babelfish

Time Limit : 6000/3000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 11   Accepted Submission(s) : 8

Problem 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

 

Source

PKU

 

解法一：Trie树

 

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

struct node{
    char wrd[15];
    node *next[26];
    node(){
        for(int i=0;i<26;i++)
            next[i]=NULL;
    }
};

node *root;

void Insert(char *str,char *wrd){
    node *loc=root,*tmp;
    for(int i=0;str[i]!='\0';i++){
        int id=str[i]-'a';
        if(loc->next[id]!=NULL){
            loc=loc->next[id];
        }else{
            tmp=new node;
            loc->next[id]=tmp;
            loc=loc->next[id];
        }
    }
    strcpy(loc->wrd,wrd);
}

int SearchTrie(char *str,char *wrd){
    node *loc=root;
    for(int i=0;str[i]!='\0';i++){
        int id=str[i]-'a';
        if(loc->next[id]!=NULL)
            loc=loc->next[id];
        else
            return 0;
    }
    strcpy(wrd,loc->wrd);
    return 1;
}

void release(node *root){
    if(root==NULL)
        return ;
    for(int i=0;i<26;i++)
        if(root->next[i]!=NULL)
            release(root->next[i]);
    delete root;
    root=NULL;
}

int main(){

    //freopen("input.txt","r",stdin);

    root=new node;
    char wrd[15],str[15];
    char s[40];
    while(gets(s)){
        if(strlen(s)==0)
            break;
        sscanf(s,"%s%s",wrd,str);
        Insert(str,wrd);
    }
    while(gets(str)){
        int flag=SearchTrie(str,wrd);
        if(flag)
            printf("%s\n",wrd);
        else
            printf("eh\n");
    }
    release(root);
    return 0;
}

 

#include<string.h>
#include<stdlib.h>
#include<stdio.h>

#define MAX 26

struct Trie{
    char *wrd;
    struct Trie *next[MAX];
};

struct Trie *Root;

void Init(){
    Root=(Trie *)malloc(sizeof(Trie));
    for(int i=0;i<MAX;i++)
        Root->next[i]=NULL;
    Root->wrd=NULL;
}

void CreateTrie(char *str,char *wrd){
    Trie *location=Root,*tmp;
    while(location!=NULL && *str!='\0'){
        int id=*str-'a';
        if(location->next[id]==NULL){
            tmp=(Trie *)malloc(sizeof(Trie));
            for(int i=0;i<MAX;i++)
                tmp->next[i]=NULL;
            tmp->wrd=NULL;
            location->next[id]=tmp;
        }
        location=location->next[id];
        str++;
    }
    if(location->wrd==NULL){
        location->wrd=new char[strlen(wrd)+1];
        strcpy(location->wrd,wrd);
    }
}

int SearchTrie(char *str,char *wrd){
    Trie *location=Root;
    while(location!=NULL && *str!='\0'){
        int id=*str-'a';
        location=location->next[id];
        str++;
    }
    if(location!=NULL && location->wrd!=NULL){
        strcpy(wrd,location->wrd);
        return 1;
    }else
        return 0;
}

int main(){
    char s[15],t[15];
    char buff[50];
    Init();
    while(gets(buff) && strcmp(buff,"")!=0){
        sscanf(buff,"%s%s",s,t);
        CreateTrie(t,s);
    }
    while(scanf("%s",s)!=EOF){
        if(SearchTrie(s,t))
            printf("%s\n",t);
        else
            printf("eh\n");
    }
    return 0;
}

解法二：hash表（待定，，转，，没大懂）

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define COLUMN 100000
#define LEN 11

//char dic[COLUMN][LEN];
struct node{
    char s[LEN];
    char sHash[LEN];
    struct node* next;
}hashNode[COLUMN];

unsigned int ELFHash(char* str)
{
    unsigned int hash = 0;
    unsigned int x = 0;

    while (*str){
        hash = (hash << 4) + (*str++);
        if ((x = hash & 0xF0000000L) != 0){
            hash ^= (x >> 24);
            hash &= ~x;
        }
    }
    return (hash & 0x7FFFFFFF);
}

void insert(char str[], char sHash[])
{
    int h = ELFHash(sHash)%COLUMN;
    struct node* p = &hashNode[h];
    while (p->next != NULL){
        p = p->next;
    }
    strcpy(p->s, str);
    strcpy(p->sHash, sHash);
    struct node* tmp = (struct node*)malloc(sizeof(struct node));
    tmp->next = NULL;
    p->next = tmp;
}

char* search(char sHash[])
{
    int h = ELFHash(sHash)%COLUMN;
    struct node* p = &hashNode[h];
    while (p->next != NULL){
        if (!strcmp(sHash, p->sHash))
            return p->s;
        p = p->next;
    }
    return NULL;
}

int main(void)
{
    char s1[LEN], s2[LEN];
    char buff[50];
    memset(hashNode, 0, sizeof(hashNode));
    while (gets(buff) && strcmp(buff, "") != 0){
        sscanf(buff, "%s%s", s1, s2);
        insert(s1, s2);
    }

    while (scanf("%s", s1) != EOF){
        char* p = search(s1);
        if (p)
            printf("%s\n", p);
        else
            printf("eh\n");
    }
    return 0;
}