首页 > 代码库 > POJ 1318 Word Amalgamation结题报告
POJ 1318 Word Amalgamation结题报告
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******
题目大意：输入几行单词作为字典，在XXXX后面输入要查找的单词，然后按顺序输出每个要查找的单词在字典中的各种顺序，每个单词后面都有星号。。。
解题思路：用一个数组来储存字典和要查找的单词，用另一个数组记录所有单词按ASC2码的排序，然后逐个枚举，找出相同的输出第一个数组中储存的单词
1 #include<stdio.h> 2 #include<string.h> 3 #include<algorithm> 4 using namespace std; 5 char str[103][7]; 6 char in[7]; 7 int cmp(char a,char b) 8 { 9 10 return strcmp(a,b)>1;11 }12 bool cmp1(char a,char b)13 {14 return a<b;15 }16 bool is_same(int len,char *in,char *a)17 {18 char out[7];19 strcpy(out,a);20 sort(out,out+len,cmp1);21 if(strcmp(out,in)==0)22 return true;23 return false;24 }25 int main()26 {27 int i=0;28 while(scanf("%s",str[i++])==1&&strcmp(str[i-1],"XXXXXX")){}29 int n=i-1;30 qsort(str,n,sizeof(str[0]),cmp);31 while(1)32 {33 scanf("%s",in);34 if(strcmp(in,"XXXXXX")==0)35 break;36 bool flag=false;37 int len=strlen(in);38 sort(in,in+len,cmp1);39 for(int i=0;i<n;i++)40 {41 if(len==strlen(str[i])&&is_same(len,in,str[i]))42 {43 printf("%s\n",str[i]);44 flag=true;45 }46 }47 if(!flag)48 printf("NOT A VALID WORD\n");49 printf("******\n");50 }51 return 0;52 }
POJ 1318 Word Amalgamation结题报告
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。