hdoj 1113 Word Amalgamation

Word Amalgamation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2854    Accepted Submission(s): 1376

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

tarpgivenscorerefundonlytrapworkearncoursepepperpartXXXXXXresconfudreaptrsettoresucXXXXXX

 

Sample Output

score******refund******parttarptrap******NOT A VALID WORD******course******
 
思路：在第一排XXXXXX的上面的字符串中寻找可以得到第一排XXXXXX的下面的字符串，字符顺序可以颠倒；先将第一排XXXXXX上面的每个字符串进行排序，每输入一个第一排XXXXXX下面的字符串也进行排序，然后依次于第一排XXXXXX上面排过序的字符串比较，存在的话按字典顺序输出原字符。
 
代码：
 
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct record{char str[100];} num[200];bool cmp(record a,record b){return strcmp(a.str,b.str)<0;}char s[100][100],cpy[100][100],a[100];int main(){int n=0,i,k,f;while(scanf("%s",s[n])&&strcmp(s[n],"XXXXXX")!=0){strcpy(cpy[n],s[n]);sort(cpy[n],cpy[n]+strlen(cpy[n]));n++;}while(scanf("%s",a)&&strcmp(a,"XXXXXX")!=0){k=0;f=0;sort(a,a+strlen(a));for(i=0;i<n;i++){if(strcmp(a,cpy[i])==0){f=1;strcpy(num[k++].str,s[i]);}}if(f==0)    {    printf("NOT A VALID WORD\n");    }    else    {    sort(num,num+k,cmp);    for(i=0;i<k;i++)    {    printf("%s\n",num[i].str);    }    }    printf("******\n");}return 0;}