HDOJ 1075 What Are You Talking About (字典树映射)

What Are You Talking About

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/204800 K (Java/Others)
Total Submission(s): 17943    Accepted Submission(s): 5836

Problem Description

Ignatius is so lucky that he met a Martian yesterday. But he didn't know the language the Martians use. The Martian gives him a history book of Mars and a dictionary when it leaves. Now Ignatius want to translate the history book into English. Can you help him?

 

Input

The problem has only one test case, the test case consists of two parts, the dictionary part and the book part. The dictionary part starts with a single line contains a string "START", this string should be ignored, then some lines follow, each line contains two strings, the first one is a word in English, the second one is the corresponding word in Martian's language. A line with a single string "END" indicates the end of the directory part, and this string should be ignored. The book part starts with a single line contains a string "START", this string should be ignored, then an article written in Martian's language. You should translate the article into English with the dictionary. If you find the word in the dictionary you should translate it and write the new word into your translation, if you can't find the word in the dictionary you do not have to translate it, and just copy the old word to your translation. Space(' '), tab('\t'), enter('\n') and all the punctuation should not be translated. A line with a single string "END" indicates the end of the book part, and that's also the end of the input. All the words are in the lowercase, and each word will contain at most 10 characters, and each line will contain at most 3000 characters.

 

Output

In this problem, you have to output the translation of the history book.

 

Sample Input

STARTfrom fiwohello difhmars riwosfearth fnnvklike fiiwjENDSTARTdifh, i'm fiwo riwosf.i fiiwj fnnvk!END

 

Sample Output

hello, i'm from mars.i like earth!
Hint
Huge input, scanf is recommended.贼烦这题，输入什么的要疯了，思路：因为要找对应，所以直接在单词的最后复制上所对应的的单词，到时候直接查找返回单词所在结点ac代码： 
#include<stdio.h>#include<string.h>#include<math.h>#include<stack>#include<iostream>#include<algorithm>#define fab(a) (a)>0?(a):(-a)#define LL long long#define MAXN 4000#define mem(x) memset(x,0,sizeof(x))#define INF 0xfffffff using namespace std;struct s{    char word[21];    s *next[26];};s *root;char str[MAXN],ss[MAXN];void create(char *str,char *ss){    int len=strlen(ss);    s *p=root,*q;    for(int i=0;i<len;i++)    {        int id=ss[i]-'a';        if(p->next[id]==NULL)        {            q=(s *)malloc(sizeof(s));            for(int j=0;j<26;j++)            q->next[j]=NULL;            strcpy(q->word,"");            p->next[id]=q;            p=p->next[id];        }        else        {            p=p->next[id];        }    }    strcpy(p->word,str);}s *find(char *sq){    int len=strlen(sq);    s *p=root;    for(int i=0;i<len;i++)    {        int id=sq[i]-'a';        //p=p->next[id];        if(p->next[id]==NULL)        return NULL;        p=p->next[id];    }    if(strcmp(p->word,""))    return p;    return NULL;}void begin(){    for(int i=0;i<26;i++)    root->next[i]=NULL;}void freetree(s *t){    if(t==NULL)    return;    for(int i=0;i<26;i++)    {        if(t->next[i]!=NULL)        freetree(t->next[i]);    }    free(t);    return;}int main(){    root=(s *)malloc(sizeof(s));    begin();    strcpy(root->word,"");    while(scanf("%s",str)!=EOF)    {        if(str[0]=='S')        continue;        if(str[0]=='E')        break;        scanf("%s",ss);        create(str,ss);    }//    if(p!=NULL)//    printf("111");//    else//    printf("000");    getchar();    while(gets(str))    {        //getchar();        if(str[0]=='S')        continue;        if(str[0]=='E')        break;        int len=strlen(str);        char ne[MAXN];        mem(ne);        for(int i=0;i<len;i++)        {            if(str[i]>='a'&&str[i]<='z')            {                int k=i;                int j=0;                while(k<len&&str[k]>='a'&&str[k]<='z')                ne[j++]=str[k++];                ne[j]='\0';                //printf("%s",ne);                s *p=find(ne);                if(p!=NULL)                printf("%s",p->word);                else                {                    for(int q=i;q<k;q++)                    printf("%c",str[q]);                }                i=k-1;                mem(ne);            }            else            printf("%c",str[i]);        }        printf("\n");    }    freetree(root);    return 0;}