Word Amalgamation(STL库的使用 + next_permutation)
来源:互联网 发布:淘宝哪家鞋子质量好 编辑:程序博客网 时间:2024/05/17 01:06
Word Amalgamation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4437 Accepted Submission(s): 2233
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:
- a dictionary, which consists of at least one and at most 100 words, one per line;
- a line containing XXXXXX, which signals the end of the dictionary;
- one or more scrambled `words’ that you must unscramble, each on a line by itself; and
- 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
//首先说下c++的一个函数 next_permutation(start,end);
start 是数组开始的位置 end 是数组结束的位置
这个函数是一个全排列的函数
#include<iostream>#include<string>#include<set>#include<algorithm>#include<stdio.h>#include<string.h>using namespace std;int main(){ char tt[8]; set<string> pp; string str; while(cin>>str&&str.compare("XXXXXX")) { pp.insert(str); } while(scanf("%s",tt)!=EOF&&strcmp(tt,"XXXXXX")) { int len = strlen(tt); int flag = 0; sort(tt,tt+len); do { str = tt; if(pp.find(str)!=pp.end()) { flag = 1; cout<<str<<endl; } }while(next_permutation(tt,tt+len)); if(!flag) { cout<<"NOT A VALID WORD"<<endl; } printf("******\n"); } return 0;}
- Word Amalgamation(STL库的使用 + next_permutation)
- POJ-1318-Word Amalgamation-STL的使用
- Word Amalgamation (stl-map)
- 使用stl的next_permutation
- poj 1318 Word Amalgamation map的使用
- POJ 1318 Word Amalgamation (字符串 STL大水)
- HDU 1113 Word Amalgamation(字符串&&STL)
- 【STL】next_permutation的原理和使用
- hdu1716 STL next_permutation函数的使用
- 【转】【STL】next_permutation的原理和使用
- 【STL】next_permutation的原理和使用
- STL算法:prev_permutation和next_permutation的使用
- 【STL】next_permutation的原理和使用
- STL算法:prev_permutation和next_permutation的使用
- [STL] next_permutation 的原理和使用
- 【STL】prev_pertutation和next_permutation的使用
- 【STL】next_permutation的原理和使用
- Word Amalgamation
- unique-binary-search-trees-ii
- 1025. 反转链表 (25)--PAT乙级
- 学习之路-Hibernate延迟加载
- Valid Parentheses--LeetCode
- Android Apk解析
- Word Amalgamation(STL库的使用 + next_permutation)
- ROS Service的使用_Python
- Innodb Cluster 入门(2) Mysql二进制日志
- SSL P2743 看电影
- 离散化-线段树-扫描线小结
- HDU 1258-Sum It Up(dfs)
- 【后缀数组】后缀排序
- android学习笔记(一)
- 《Linux程序设计》 -> 《Linux高级程序设计》 -> 《Unix环境高级编程》