Word Amalgamation

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
******
 
分析:题意是给你一个单词字典,每次询问一串字符,问能否组成字典里的单词,并按字典序输出。这题开始尝试暴力查找,但没注意到要按字典序,后来尝试用set,但第三个样例出错了,怎么也想不通,最后换成map,把字典里的每个单词里的字符排序并与原单词对应,然后查找时排个序再比较就很方便了。
最后说一句,map大法好~~~
 
 #include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
using namespace std;
#define INF -100000
typedef long long ll;
const int maxn=;
string words,tmp,query;
map<string,string> ans;
int main()
{
// freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
while(cin>>words){
if(words=="XXXXXX") break;
tmp=words;
sort(words.begin(),words.end());
ans[tmp]=words;
}
while(cin>>query){
if(query=="XXXXXX") break;
int flag=;
sort(query.begin(),query.end());
for(map<string,string>::iterator it=ans.begin();it!=ans.end();it++)
if(it->second==query){
cout<<it->first<<endl;
flag=;
}
if(!flag) printf("NOT A VALID WORD\n");
printf("******\n");
}
return ;
}

HDU-1113(map的运用)的更多相关文章

  1. HDU 1113 Word Amalgamation (map 容器 + string容器)

    http://acm.hdu.edu.cn/showproblem.php?pid=1113 Problem Description In millions of newspapers across ...

  2. HDOJ/HDU 1113 Word Amalgamation(字典顺序~Map)

    Problem Description In millions of newspapers across the United States there is a word game called J ...

  3. hdu 1113 Word Amalgamation

    原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1113 字符串简单题: stl水过 如下: #include<algorithm> #inc ...

  4. hdu - 1113 Word Amalgamation (stl)

    http://acm.hdu.edu.cn/showproblem.php?pid=1113 给定一个字典,然后每次输入一个字符串问字典中是否有单词与给定的字符串的所有字母一样(顺序可以打乱),按字典 ...

  5. hdu 1113 Word Amalgamation 解题报告

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1113 题意:输入一个字典,然后再输入若干单词(每行中,1 <= 单词数 <= 100,并且 ...

  6. hdu 1075 (map)

    http://acm.hdu.edu.cn/showproblem.php?pid=1075 What Are You Talking About Time Limit: 10000/5000 MS ...

  7. hdu 1247 map的使用

    http://acm.hdu.edu.cn/showproblem.php?pid=1247 Hat’s Words Time Limit: 2000/1000 MS (Java/Others)    ...

  8. hdu 4941 map的使用

    http://acm.hdu.edu.cn/showproblem.php?pid=4941 给定N,M和K,表示在一个N*M的棋盘上有K个棋子,给出K个棋子的位置和值,然后是Q次操作,对应的是: 1 ...

  9. hdu 2112 HDU Today(map与dijkstra的结合使用)

    HDU Today Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total ...

  10. HDU 5167(map + 暴力)

    题意:给出一个数n,问n能否是斐波那契数列中数的乘积 先刷选 斐波那契数列,然后就枚举 #include <cstdio> #include <cstring> #includ ...

随机推荐

  1. Android Density(密度)

    1.  什么是density 引用 1) density density表示每英寸有多少个显示点(逻辑值),它的单位是dpi:dot per inch,通常屏幕大时,density就大,屏幕小时,de ...

  2. PLSQL Developer Debug

    如果要查看存储过程或者函数的执行过程,可以用debug的模式.PLSQL Developer提供了debug功能,以函数为例: 1. 找到你要debug的函数,然后右击—>选择“Add debu ...

  3. 关于Json传递的日期/Date(数字)/解析

    在将DateTime类型的数据Json后传到前台展示,出现如下效果 ,在客户端如何解析呢?在jquery easyui 的字段中加一个格式化的函数调用.   { field: 'CreateTime' ...

  4. bzoj3875

    悲伤地回忆,当初写了一个作死的算法爆零了为什么不好好写暴力呢……显然设w[i]表示彻底干掉这个怪物的代价注意发现这里的转移具有后效性,但是干掉每个怪物的最优值是一定的我们用spfa来转移,详见那篇sp ...

  5. 【转】virtualbox 4.08安装虚机Ubuntu11.04增强功能失败解决方法

    原文网址:http://fuliang.iteye.com/blog/1102998 在笔记本安装Ubuntu11.04增强功能失败 引用 fuliang@fuliang-VirtualBox:~$ ...

  6. CSS实现Div透明,而显示在上面的文字不透明,但也可看到显示在下面的图片内容

    CSS实现Div透明,而显示在上面的文字不透明,但也可看到显示在下面的图片内容,DiV透明其实挺简单,主要是为background定义opacity属性,一般这个是最大值是1,数值越接近1,则越不透明 ...

  7. Android4.0 -- UI控件之 Menu 菜单的的使用(二)

    上一讲我们讲解了android中在代码或者xml文件中定义菜单,这一讲我们继续来讲解一下定义菜单的其他方式:创建上下文的菜单.查看API文档 Menus :Creating Contextual Me ...

  8. HDU 5938 Four Operations 【贪心】(2016年中国大学生程序设计竞赛(杭州))

    Four Operations Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)T ...

  9. [Boost::Polygon]多边形相减得到新的多边形序列

    #include <iostream> #include <boost/polygon/polygon.hpp> #include <cassert> namesp ...

  10. Hadoop之hive安装过程以及运行常见问题

    Hive简介 1.数据仓库工具 2.支持一种与Sql类似的语言HiveQL 3.可以看成是从Sql到MapReduce的映射器 4.提供shall.Jdbc/odbc.Thrift.Web等接口 Hi ...