UVa 642 Word Amalgamation (查字典&字符串排序)_uva642 word amalgamation-优快云博客

本文链接：https://blog.youkuaiyun.com/synapse7/article/details/12114343

本文介绍了一个程序设计挑战，旨在从一个包含字母的字谜中解码并重组单词。通过输入一个单词字典、一组需要解码的乱序单词以及结束标志，程序能够识别所有可以由乱序单词重新排列形成的字典中的单词，并以字典序输出这些单词。如果不存在匹配的单词，则输出提示信息。示例输入包括一个单词集合和一组乱序单词，程序演示了如何识别并输出所有可能的组合。

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642 - Word Amalgamation

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=onlinejudge&page=show_problem&problem=583

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 file 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 uppercaseX'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
******

统统排序。

完整代码：

/*0.015s*/

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

char word[110][10], sorted[110][10], s[10];
char *pstr[110];

bool cmp_string(const char *a, const char *b)
{
	return strcmp(a, b) < 0;
}

int main(void)
{
	int n = 0;
	while (gets(word[n]), word[n][0] != 'X')
	{
		pstr[n] = word[n];
		++n;
	}
	///交换两个字符串数组？干脆交换指针吧。
	sort(pstr, pstr + n, cmp_string);
	///注意在排序之后，word的顺序并未改变，但sorted是字典序
	for (int i = 0; i < n; i++)
	{
		strcpy(sorted[i], pstr[i]);
		sort(sorted[i], sorted[i] + strlen(sorted[i]));
	}
	while (gets(s), s[0] != 'X')
	{
		sort(s, s + strlen(s));
		bool found = false;
		for (int i = 0; i < n; i++)
			if (strcmp(sorted[i], s) == 0)
			{
				found = true;
				puts(pstr[i]);
			}
		if (!found)
			puts("NOT A VALID WORD");
		puts("******");
	}
	return 0;
}