HDU1075-What Are You Talking About

What Are You Talking About

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 102400/204800 K (Java/Others)

Total Submission(s): 20751    Accepted Submission(s): 6881

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.

 

题目大意：

相当于就是给了原文和对应的译文，然后给出例句，输出真正想要说的内容。都是以START开始和END结束。

解题思路：

用map做的话很简单，要是需要的话后面在补上。这里我是用字典树做的，缺点就是容易超内存。麻烦的地方时就是树需要自己建。下面的代码里面注释写的已经很详细了，所以就不再说了。

源代码：

<span style="font-size:18px;">

#include<iostream>
#include<algorithm>
#include<string>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;

const int MAXN = 26;
const int MAXM = 15;
const int MAXS = 3005;

struct Trie//名字不用起多，按书上的写太累赘
{
	char s[MAXM];
	Trie *next[MAXN];
};
Trie *root;//紧随其后，声明一个

void init()
{
	root = (Trie*)malloc(sizeof(Trie));
	strcpy(root->s, "");
	for (int i = 0; i<MAXN; i++)
	{
		root->next[i] = NULL;
	}
}

void CreateTire(char str[], char ss[])//不许要传整棵树了
{
	//这里需要p来对roor进行操作，因为会有迭代过程
	//如果root参与运算的话，root=root->next[i],root的值就被改变了
	Trie *p = root, *temp;
	int len = strlen(ss);
	for (int i = 0; i<len; i++)
	{
		int id = ss[i] - 'a';
		if (p->next[id] == NULL)//说明还没有访问过
		{
			//对临时temp进行初始话，同root的初始化一样
			temp = (Trie*)malloc(sizeof(Trie));
			strcpy(temp->s, "");
			for (int j = 0; j<MAXN; j++)
			{
				temp->next[j] = NULL;
			}
			p->next[id] = temp;
			p = p->next[id];
		}
		else//访问过继续向下走
		{
			p = p->next[id];
		}//到最后了，把需要翻译的字符串拷贝进来
		if (i == len - 1)
		{
			strcpy(p->s, str);
		}
	}
	//建树成功
}

string FindTrie(char ss[])//写string,不要写成char*,会返回地址
{
	string ans = "";
	Trie *p = root;
	int len = strlen(ss);
	for (int i = 0; i<len; i++)
	{
		int id = ss[i] - 'a';
		if (p->next[id] == NULL)
		{
			ans = "";//没有的话就把空串拷贝进去
			return ans;
		}
		else
		{
			p = p->next[id];//有的话继续向下
		}
	}
	//假设ss对应的没有保存，因为ss可能是另外一个的前缀，这个时候p->next[id]也不为空
	//但是拷贝进来的还是空串，不用担心
	ans = (string)p->s;
	return ans;
}

void getData()
{
	char s[MAXM];
	scanf("%s", s);//先把start读进来
	while (~scanf("%s", s))
	{
		if (s[0] == 'E')//读到end退出
			break;
		char ss[MAXM];
		scanf("%s", ss);
		CreateTire(s, ss);
	}
}

void slove()
{
	string s;//因为整行读取还是string方便
	cin >> s;
	getline(cin, s);
	while (getline(cin, s))
	{
		if (s[0] == 'E')
			break;
		int st;//开始结束的下标
		st = 0;
		string res = "";
		s += ".";//末尾加一个不是小写字母的字符方便处理
		int len = s.length();
		for (int i = 0; i<len; i++)
		{
			if (s[i]>'z'||s[i]<'a')//不属于小写字母，逻辑或的关系
			{
				char temp[MAXM];
				int index = 0;
				for (int j = st; j<i; j++)
				{
					temp[index] = s[j];
					index++;
				}
				temp[index] = '\0';//'\0'结尾
				string ss = FindTrie(temp);
				if (ss=="")//是空串
				{
					res += (string)temp;//保留原值追加
				}
				else
				{
					res += ss;
				}
				//寻找下一个小写字母开头的位置赋值st
				st = i;
				while (s[i]>'z'||s[i]<'a'&&i<len-1)
				{
					i++;
				}
				index = 0;
				//把中间不是小写字符的字符串原值保留
				for (int j = st; j < i; j++)
				{
					temp[index] = s[j];
					index++;
				}
				temp[index] = '\0';
				res += (string)temp;
				st = i;

			}
		}
		cout << res << endl;
	}
	delete(root);
}
int main()
{
	init();
	getData();
	slove();
	return 0;
}

</span>