hdu1075（trie树）

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

     
     
     
     

      
      
      
      START
from fiwo
hello difh
mars riwosf
earth fnnvk
like fiiwj
END
START
difh, i'm fiwo riwosf.
i fiiwj fnnvk!
END 
     
     
     
     

 

Sample Output

     
     
     
     

      
      
      
      hello, i'm from mars.
i like earth!


         
         
     
     
     
     

Hint

 Huge input, scanf is recommended. 

输入数据量比较大，其实也就是最基础的trie树啦，不过！这题如果是你想象中辣么简单就大错特错啦，，，，，，，，，5555，硬生生被坑在了数据上（此题如果句末有标点符号，才能对应输出！！！！！！！否则根本没输出555555555555，就因为这个wa了十几次，还有为这题苦思冥想觉得自己写的很对结果就是wa的骚年吗？！谨记我的教训啊555555555）

#include <iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int NODE = 1e7+10,CH = 26;

int ch[NODE][CH],sz,val[NODE];
char a1[1000001][15],a2[15];
    /**
    ch[u][c]:   节点u指向的c儿子的边
    val[u]:     节点u的值
    sz:         trie树的size
    */
int idx(char c)
{return c-'a';}
int node()
{
        /**
        新建(初始化)节点
        **/
        memset(ch[sz],0,sizeof(ch[sz]));
        ///将所有儿子置为空
        val[sz]=0;
        return sz++;
}
void init()
{
        sz=0;
        ///trie树根节点为下标0的节点
        node();
}
void insert(char *s,int v)
{
        int u=0;
        for(;*s;s++)
        {
            int c=idx(*s);
            ///如果节点不存在 新建节点 并把值赋给当前节点的c儿子边
            if(!ch[u][c])
                ch[u][c]=node();
            ///继续移动
            u=ch[u][c];
        }
        ///在末尾节点记录信息
        val[u]=v;
}

int find(char *s)
{
    int u=0;
    for(;*s;s++)
    {
        int c=idx(*s);
        ///如果u节点没有c儿子  结束
            if(!ch[u][c])
            return 0;
            u=ch[u][c];
    }
        return val[u];
}

int zm(char x)
{
    if(x>='a'&&x<='z')
    return 1;
    return 0;
}

int main()
{
    char start[7],chu[15];
    scanf("%s",start);
    int o=0;
    while(~scanf("%s",a1[o]))
    {
        if(strcmp(a1[o],"END")==0)
        break;
        scanf("%s",a2);
        if(o==0)
        strcpy(chu,a2);
        insert(a2,o++);
    }

    char b1[3005];
    scanf("%s",start);
    getchar();
    while(gets(b1))
    {
        if(strcmp(b1,"END")==0)
        break;
        int i,l=strlen(b1),flag=0,k=0;
        char t[15];
        for(i=0;i<l;i++)
        if(flag==0&&zm(b1[i])==1)
            t[k++]=b1[i],flag=1;
        else if(flag==1&&zm(b1[i])==1)
            t[k++]=b1[i];
        else if(flag==1&&zm(b1[i])==0)
        {
            t[k]='\0';
            if(strcmp(chu,t)==0)
            {
                 printf("%s",a1[0]);memset(t,0,sizeof(t));
            k=0,flag=0;
           printf("%c",b1[i]);continue;
           }
            if(find(t)==0)
            printf("%s",t);
            else
            printf("%s",a1[find(t)]);
            memset(t,0,sizeof(t));
            k=0,flag=0;
            printf("%c",b1[i]);
        }
        else if(flag==0&&zm(b1[i])==0)
        printf("%c",b1[i]);
        printf("\n");
    }
    return 0;
}