POJ 1318 Word Amalgamation (字符串 STL大水)
2015-07-04 00:34
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Word Amalgamation
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
Sample Output
Source
Mid-Central USA 1998
题目链接:http://poj.org/problem?id=1318
题目大意:给一个字典,"XXXXXX"结束,再输入一些打乱的字母,"XXXXXX"结束,现在重排这些字母,问重排后的单词是否出现在字典中,出现就输出,可能有不同多个
题目分析:因为单词长度最大为6,直接next_permutation,字典用set存,0ms
代码:
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 8665 | Accepted: 4172 |
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
题目链接:http://poj.org/problem?id=1318
题目大意:给一个字典,"XXXXXX"结束,再输入一些打乱的字母,"XXXXXX"结束,现在重排这些字母,问重排后的单词是否出现在字典中,出现就输出,可能有不同多个
题目分析:因为单词长度最大为6,直接next_permutation,字典用set存,0ms
代码:
#include <iostream> #include <cstdio> #include <string> #include <algorithm> #include <set> using namespace std; int main() { int cnt = 0; string s; set <string> st; set <string> :: iterator it; while(cin >> s) { if(s == "XXXXXX") { cnt ++; continue; } if(cnt == 2) break; if(cnt == 0) st.insert(s); else if(cnt == 1) { bool flag = false; sort(s.begin(), s.end()); do { it = st.find(s); if(it != st.end()) { flag = true; cout << s << endl; } }while(next_permutation(s.begin(), s.end())); if(!flag) printf("NOT A VALID WORD\n"); printf("******\n"); } } }
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