HDU 1075 What Are You Talking About（Trie的应用）

What Are You Talking About

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/204800 K (Java/Others)
Total Submission(s): 20680    Accepted Submission(s): 6852

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!

Hint

Huge input, scanf is recommended.

 

题目链接：HDU 1075

结构体Trie中用flag=1表示当前点是否是某个单词的结尾处（防止出现单词a为单词b的前缀但是却被b覆盖的情况，比如diff对应bbb，但是此时diff的前缀dif显然是不存在的），若为结尾，则flag赋值为1，然后把对应的翻译单词赋值给这个节点的s[]。

查询的是否看查完之后的flag是否为1，如果为1则存在该单词，否则flag为0或根本查不到，则返回NULL

一开始智障了把每个点flag都设为1狂WA……

代码：

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <sstream>
#include <cstring>
#include <bitset>
#include <string>
#include <deque>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=26;
const int M=100;
struct Trie
{
    Trie *nxt[N];
    int flag;
    char s[M];
    Trie()
    {
        CLR(nxt,0);
        CLR(s,0);
        flag=0;
    }
};
Trie *L=new Trie();
void update(char s[],char earth[])
{
    int len=strlen(s);
    int indx;
    Trie *cur=L;
    for (int i=0; i<len; ++i)
    {
        indx=s[i]-'a';
        if(cur->nxt[indx])
            cur=cur->nxt[indx];
        else
        {
            Trie *one=new Trie();
            cur->nxt[indx]=one;
            cur=one;
        }
    }
    strcpy(cur->s,earth);
    cur->flag=1;
}
char* Find(char s[])
{
    int len=strlen(s);
    int indx;
    Trie *cur=L;
    for (int i=0; i<len; ++i)
    {
        indx=s[i]-'a';
        if(cur->nxt[indx]==NULL)
            return NULL;
        cur=cur->nxt[indx];
    }
    return cur->flag?cur->s:NULL;
}
char from[M],to[M];
char tot[1000010];
char temp[M];
int main(void)
{
    int i;
    scanf("%s",from);
    getchar();
    while (~scanf("%s",to)&&strcmp(to,"END"))
    {
        scanf("%s",from);
        update(from,to);
    }
    getchar();

    gets(from);
    while (gets(tot)&&strcmp(tot,"END"))
    {
        int len=strlen(tot);
        int k=0;
        CLR(temp,0);
        for (i=0; i<len; ++i)
        {
            if(islower(tot[i]))
                temp[k++]=tot[i];
            else
            {
                char *one=Find(temp);
                if(k)
                    printf("%s",one==NULL?temp:one);
                putchar(tot[i]);
                CLR(temp,0);
                k=0;
            }
        }
        if(k)
        {
            char *one=Find(temp);
            printf("%s",one==NULL?temp:one);
        }
        putchar('\n');
    }
    return 0;
}