首页 > 代码库 > POJ 1318
POJ 1318
Word Amalgamation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4340 Accepted Submission(s): 2187
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.
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
******
先给定一个字典,再给一堆字符,每一串字符随机组合成字典中的单词,若可以组合成则单词字典序输出,否则输出无效的单词,每一个单词查找完后输出一行******
容器+排序
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <vector> #include <queue> #include <cstdlib> #include <iomanip> #include <cmath> #include <ctime> #include <map> #include <set> using namespace std; #define lowbit(x) (x&(-x)) #define max(x,y) (x>y?x:y) #define min(x,y) (x<y?x:y) #define MAX 100000000000000000 #define MOD 1000000007 #define pi acos(-1.0) #define ei exp(1) #define PI 3.141592653589793238462 #define INF 0x3f3f3f3f3f #define mem(a) (memset(a,0,sizeof(a))) typedef long long ll; map<string,int>m; set<string>s[105]; set<string>::iterator it; string str,rstr,name="XXXXXX"; int main() { int tot=1; while(cin>>str) { if(str==name) break; rstr=str; sort(rstr.begin(),rstr.end()); if(m[rstr]==0) m[rstr]=tot++; s[m[rstr]].insert(str); } while(cin>>str) { if(str==name)break; sort(str.begin(),str.end()); if(m[str]==0) printf("NOT A VALID WORD\n******\n"); else { for(it=s[m[str]].begin();it!=s[m[str]].end();it++) cout<<*it<<endl; printf("******\n"); } } return 0; }
POJ 1318
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。