hdu1113 Word Amalgamation（详解--map和string的运用）

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题目链接：http://acm.hdu.edu.cn/showproblem.php?

pid=1113

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Word Amalgamation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2496    Accepted Submission(s): 1198

Problem Description

In millions of newspapers across the United States there is a word game called Jumble. The object of this game is to solve a riddle, but in order to find the letters that appear in the answer it is necessary to unscramble four words. Your task is to write a program that can unscramble words. 

 

Input

The input contains four parts: 

1. a dictionary, which consists of at least one and at most 100 words, one per line; 
2. a line containing XXXXXX, which signals the end of the dictionary; 
3. one or more scrambled `words' that you must unscramble, each on a line by itself; and 
4. another line containing XXXXXX, which signals the end of the file.

All words, including both dictionary words and scrambled words, consist only of lowercase English letters and will be at least one and at most six characters long. (Note that the sentinel XXXXXX contains uppercase X's.) The dictionary is not necessarily in sorted order, but each word in the dictionary is unique. 

 

Output

For each scrambled word in the input, output an alphabetical list of all dictionary words that can be formed by rearranging the letters in the scrambled word. Each word in this list must appear on a line by itself. If the list is empty (because no dictionary words can be formed), output the line ``NOT A VALID WORD" instead. In either case, output a line containing six asterisks to signal the end of the list.

 

Sample Input

tarp given score refund only trap work earn course pepper part XXXXXX resco nfudre aptr sett oresuc XXXXXX

 

Sample Output

score ****** refund ****** part tarp trap ****** NOT A VALID WORD ****** course ******

 

Source

Mid-Central USA 1998

 

Recommend

Eddy

题意：先给你一些单词作为字典，在给一系列的单词查找字典中是否有这些单词（注意查找的单词，一个单词中的字母顺序是能够变得，也就是说单词之间仅仅要字母是一样的不用考虑顺序是否一样都要输出）；

思路：用map和string便非常easy解决，先把字典存入map里，在逐一查找就OK。当然查找的时候须要一点小小的操作，详见代码解释；

map的具体使用方法：http://blog.csdn.net/u012860063/article/details/24435211

代码例如以下：

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

int main()
{
	map<string,string>m;//定义map的变量和值都为string类型 
	string t, s;
	while(cin >> s)
	{
		if( s == "XXXXXX") 
		{	//注：string类型的是能够直接在字符串之间用"="或"=="进行赋值或推断的
			break;
		}
		t = s;
		sort(s.begin(),s.end());//直接对string类型的字符串用begin()和end()进行排序 
		m[t] = s;
	}
	while(cin >> s)
	{
		if( s == "XXXXXX")
		{
			break;
		}
		int flag = 0;//用于记录字典里是否有要查询的单词 
		sort(s.begin(),s.end());//直接对string类型的字符串用begin()和end()进行排序 
		map<string,string>::iterator it;
		for(it = m.begin(); it != m.end(); it++)
		{
			if(it->second == s)//it->second 为值，也就是map里第二个string的值 
			{
				cout<<it->first<<endl;//it->first 为索引键值,也就是map里第一个string的值 
				flag = 1;
			}
		}
		if(flag == 0)
		cout << "NOT A VALID WORD"<<endl; //和以下凝视掉的等效 
		cout << "******"<<endl;
	//	cout << "NOT A VALID WORD
"<<; 
	//	cout << "******
"<<;
	}
	return 0;
}