HDU 1075 What Are You Talking About(字典树学习题)

链表：

http://acm.hdu.edu.cn/showproblem.php?pid=1075

题目:

What Are You Talking About

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/204800 K (Java/Others)
Total Submission(s): 7939    Accepted Submission(s): 2470

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

   
   
   
   

    
    
    
    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! 
    
    
    
    

   
   
   
   

分析与总结：

弱逼第一次写字典树 = =|| ， 找了几分资料研究了后，自己敲起来， 结果因为数组开太小WA了，改了大了之后就科学地AC了。

先写了一个动态分配内存版本，用C++提交，用了250MS。

然后又写了一个静态内存开辟的版本， 用了187MS，  看来静态的确实比较快。

代码：

1. 动态分配内存版

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cctype>
using namespace std;

const int KIND = 26;

struct node{
    int  pos;
    bool isword;
    node *next[KIND];
    node(){
        isword=false;
        memset(next, 0, sizeof(next));
    }
};

void insert(node *root, char *str, int pos){
    for(char *p=str; *p; ++p){
        int son = *p-'a';
        if(root->next[son]==NULL){
            root->next[son] = new node;
        }
        root = root->next[son];
    }
    root->isword = true;
    root->pos    = pos;
}
int search(node *root, char *str){
    for(char *p=str; *p; ++p){
        int son = *p-'a';
        if(root->next[son]==NULL)
            return -1;
        root = root->next[son];
    }
    if(root->isword)return root->pos;
    return -1;
}

char word[500010][12];
int pos;

int main(){ 
    char str[3005], str2[3005];
    pos=0;
    gets(str);
    node *root = new node;
    while(scanf("%s %s",str,str2)){
        if(str[0]=='E')break;
        strcpy(word[pos], str);
        insert(root, str2, pos++); 
    }
    gets(str);
    while(gets(str)){
        if(str[0]=='E')break;
        bool flag=false;
        int k=0;
        for(char *p=str; *p; ++p){
            if(!flag&&!isalpha(*p)){
                printf("%c",*p);
            }
            else if(flag && !isalpha(*p)){
                flag=false;
                int t=search(root, str2);
                if(t==-1) printf("%s%c",str2,*p);
                else printf("%s%c",word[t],*p);
            }
            else if(flag){
                str2[k++]=*p;
            }
            else if(!flag&&isalpha(*p)){
                flag=true;
                k=0; 
                memset(str2, 0, sizeof(str2));
                str2[k++]=*p;
            }
        } 
        puts("");
    }
    return 0;
}

2. 静态开辟内存空间版

用了个小技巧， 单独写了一个"new node"的newNode()函数，这样只需要把上面代码分配空间部分都改为用这个函数代替就OK了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cctype>
using namespace std;

const int KIND = 26;

struct node{
    int  pos;
    bool isword;
    node *next[KIND];
    void init(){
        isword=false;
        memset(next, 0, sizeof(next));
    }
}arr[1000000];
int cntNode;

inline node* newNode(){
    arr[cntNode].init();
    return &arr[cntNode++];
}

void insert(node *root, char *str, int pos){
    for(char *p=str; *p; ++p){
        int son = *p-'a';
        if(root->next[son]==NULL){
            root->next[son] = newNode();
        }
        root = root->next[son];
    }
    root->isword = true;
    root->pos    = pos;
}
int search(node *root, char *str){
    for(char *p=str; *p; ++p){
        int son = *p-'a';
        if(root->next[son]==NULL)
            return -1;
        root = root->next[son];
    }
    if(root->isword)return root->pos;
    return -1;
}

char word[500010][12];
int pos;

int main(){ 
    char str[3005], str2[3005];
    pos=0;
    cntNode=0;

    gets(str);
    node *root = newNode();

    while(scanf("%s %s",str,str2)){
        if(str[0]=='E')break;
        strcpy(word[pos], str);
        insert(root, str2, pos++); 
    }
    gets(str);
    while(gets(str)){
        if(str[0]=='E')break;
        bool flag=false;
        int k=0;
        for(char *p=str; *p; ++p){
            if(!flag&&!isalpha(*p)){
                printf("%c",*p);
            }
            else if(flag && !isalpha(*p)){
                flag=false;
                int t=search(root, str2);
                if(t==-1) printf("%s%c",str2,*p);
                else printf("%s%c",word[t],*p);
            }
            else if(flag){
                str2[k++]=*p;
            }
            else if(!flag&&isalpha(*p)){
                flag=true;
                k=0; 
                memset(str2, 0, sizeof(str2));
                str2[k++]=*p;
            }
        } 
        puts("");
    }
    return 0;
}

  
   
 ——  生命的意义，在于赋予它意义。
   
    
  
  
  

     
   
   
             
    
  
  
  
    
  
  
  

     
   
   
        原创 http://blog.csdn.net/shuangde800 ， By   D_Double  (转载请标明)