POJ 1035 Spell checker
来源:互联网 发布:淘宝卖家在闲鱼卖东西 编辑:程序博客网 时间:2024/06/10 00:56
字符数组开小了wa好多次。。。。pe一次 哎水题
Spell checker
Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 22862 Accepted: 8314
Description
You, as a member of a development team for a new spell checking program, are to write a module that will check the correctness of given words using a known dictionary of all correct words in all their forms.
If the word is absent in the dictionary then it can be replaced by correct words (from the dictionary) that can be obtained by one of the following operations:
?deleting of one letter from the word;
?replacing of one letter in the word with an arbitrary letter;
?inserting of one arbitrary letter into the word.
Your task is to write the program that will find all possible replacements from the dictionary for every given word.
If the word is absent in the dictionary then it can be replaced by correct words (from the dictionary) that can be obtained by one of the following operations:
?deleting of one letter from the word;
?replacing of one letter in the word with an arbitrary letter;
?inserting of one arbitrary letter into the word.
Your task is to write the program that will find all possible replacements from the dictionary for every given word.
Input
The first part of the input file contains all words from the dictionary. Each word occupies its own line. This part is finished by the single character '#' on a separate line. All words are different. There will be at most 10000 words in the dictionary.
The next part of the file contains all words that are to be checked. Each word occupies its own line. This part is also finished by the single character '#' on a separate line. There will be at most 50 words that are to be checked.
All words in the input file (words from the dictionary and words to be checked) consist only of small alphabetic characters and each one contains 15 characters at most.
The next part of the file contains all words that are to be checked. Each word occupies its own line. This part is also finished by the single character '#' on a separate line. There will be at most 50 words that are to be checked.
All words in the input file (words from the dictionary and words to be checked) consist only of small alphabetic characters and each one contains 15 characters at most.
Output
Write to the output file exactly one line for every checked word in the order of their appearance in the second part of the input file. If the word is correct (i.e. it exists in the dictionary) write the message: " is correct". If the word is not correct then write this word first, then write the character ':' (colon), and after a single space write all its possible replacements, separated by spaces. The replacements should be written in the order of their appearance in the dictionary (in the first part of the input file). If there are no replacements for this word then the line feed should immediately follow the colon.
Sample Input
iishashavebemymorecontestmetooifaward#meawaremcontesthavooorifimre#
Sample Output
me is correctaware: awardm: i my mecontest is correcthav: has haveoo: tooor:i is correctfi: imre: more me
Source
Northeastern Europe 1998
[Submit] [GoACcode:
#include <cstdio>#include <cstring>#include <stdlib.h>#include <iostream>#include <algorithm>#define maxn 10000+5using namespace std;struct Note{ char word[20];}my[maxn];int n=0;char t[20];char txt[maxn][20];int tlen[maxn];Note *pos;int cmp(const void *a,const void *b){ return strcmp(((Note*)a)->word,((Note*)b)->word);};int bcmp(const void *a,const void *b){ return strcmp(((char*)a),((Note*)b)->word);};bool fun1(char *a,char *b,int l){ int cnt=0; for(int i=0;i<l;++i){ if(a[i]==b[i]) cnt++; } if(cnt+1==l)return true; return false;};bool fun2(char *a,char *b){ int flag=0; while(*a){ if(*a!=*b){ b++; flag++; if(flag>1) return false; } else { a++; b++; } } return true;};int main(){ int n=0; while(scanf("%s",txt[n])!=EOF&&strcmp("#",txt[n])){ strcpy(my[n].word,txt[n]); tlen[n]=strlen(txt[n]); n++; } qsort(my,n,sizeof(Note),cmp); while(scanf("%s",t)!=EOF&&strcmp("#",t)){ pos=(Note*)bsearch(t,my,n,sizeof(Note),bcmp); if(pos)printf("%s is correct\n",pos->word); else{ printf("%s:",t); int len=strlen(t); for(int i=0;i<n;++i){ if(len+1==tlen[i]){ if(fun2(t,txt[i])) printf(" %s",txt[i]); } else if(len==tlen[i]){ if(fun1(t,txt[i],len)) printf(" %s",txt[i]); } else if(len-1==tlen[i]){ if(fun2(txt[i],t)) printf(" %s",txt[i]); } } putchar('\n'); } } return 0;}
0 0
- poj-1035-Spell checker
- poj 1035 Spell checker
- Poj 1035 --Spell checker
- Spell checker--POJ 1035
- poj 1035 Spell checker
- poj 1035 Spell checker
- POJ-1035-Spell checker
- poj 1035 Spell Checker
- POJ 1035 Spell checker
- poj 1035 Spell checker
- poj 1035 Spell checker
- poj 1035 Spell checker
- POJ 1035 Spell checker
- POJ 1035 Spell checker
- poj 1035 Spell checker
- POJ 1035 Spell checker
- poj 1035 Spell checker
- POJ 1035 Spell checker
- 获取自定义Calendar中的数据
- 学习笔记之空指针。 新手。
- CLRS 7.2快速排序的性能
- Mac下和Windows下UnrealEngine 4体验对比
- 关于nil和 null和NSNull的区别及相关问题
- POJ 1035 Spell checker
- arm wifi驱动编译及无线wifi网络管理
- 面向服务与微服务架构
- Note For Linux By Jes(1)-Linux 的文件权限与目录配置
- 正则表达式(三) 表达式助手
- Hdu1233 最小生成树_还是畅通工程
- error C4430: 缺少类型说明符 - 假定为 int。注意: C++ 不支持默认 int
- 两分钟学会在GitHub托管代码
- Linux LVS-DR模型实战演示