HDU 1483 Automatic Correction of Misspellings （字符串大模拟）

Description

Some text editors offer a feature to correct words which seem to be written incorrectly. In this problem you are asked to implement a simple Automatic Correction of Misspellings (ACM). 

ACM takes care of the following misspellings of words: 

1.One letter is missing (e.g., letter is written leter) or too much (e.g., letter is written lettter). 

2.One letter is wrong (e.g., letter is written ketter) 

3.The order of two adjacent letters is wrong (e.g., letter is written lettre) 

ACM is based on a dictionary of known words. When a text contains a word which is not in the dictionary, ACM will try to replace it by a similar word of the dictionary. Two words are similar if we can transform one word into the other by doing exactly one of
the misspellings listed above. An unknown word is left unchanged if there is no similar word in the dictionary. 

Input

The first line of the input file will give the number n of words in the dictionary (n ≤ 10000). The next n lines contain the dictionary words. The following line contains an integer q ≤ 1000, the number of query words. The next q lines contain the query words.
You may assume that each word in the input consists of 1 to 25 lower case letters ('a' to 'z'). 

Output

For each query word, print one line with the query word followed by one of the following possibilities: 

1.is correct, if the word occurs in the dictionary. 

2.is a misspelling of <x>, where <x> is a word of the dictionary similar to the query word, and the query word is not in the dictionary. In the case that there are several possibilities, select the word from the dictionary which appeared earlier in the input. 

3.is unknown, if cases 1 and 2 do not apply. 

Sample Input

10
this
is
a
dictionary
that
we
will
use
for
us
6
su
as
the
dictonary
us
willl

Sample Output

su is a misspelling of us
as is a misspelling of is
the is unknown
dictonary is a misspelling of dictionary
us is correct
willl is a misspelling of will

这个题是说先给你一个n，然后输入n个字符串（当作字典）。然后再输入q，再输入q个字符串，，最后分别看这q个字符串和这个字典的关系。

1.字典中字符串是比该字串长度少1的子串，或者反过来也可以。

2.相邻两个字符相反。

3.只允许有一个字符出现错误。

4.该字符串在字典中。

5.如果不符合以上四个条件的，则说明未出现过。

6.当有多个符合条件的字符串时，优先出现的优先选择。

就是暴力+模拟。

#include <iostream>
#include <cstdio>
#include <string>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <map>
#include <queue>
#include <stack>
#include <functional>
#define INF 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a));
#define For(a,b) for(int i = a;i<b;i++)
#define LL long long
#define MAX_N 100010
using namespace std;
char z[10005][100],x[1005][100];
int n,k;
inline bool compare(int cur)
{
for(int i = 0; i<n; i++)
{
if(!strcmp(x[cur],z[i]))
{
return true;
}
}
return false;
}
inline bool compare1(int cur,int *pos)//zichuan
{
for(int i = 0; i<n; i++)
{
int ans = 0;
*pos = i;
int sum = 0,sum1 = 0;
int len = strlen(x[cur]);
int len1 = strlen(z[i]);
if(abs(len - len1) == 1)
{
if(len < len1)
{
int xx = 0;
sum = 0;
for(int j = 0; j<len1; j++)
{
if(x[cur][xx] == z[i][j])
xx ++;
else
sum ++;
}
}
else
{
int zz = 0;
sum1 = 0;
for(int j = 0; j<len; j++)
{
if(x[cur][j] == z[i][zz])
zz ++;
else
sum1 ++;
}
}
if(sum == 1 || sum1 == 1)
{
//cout<<"zichuan"<<endl;
return true;
}
}
}
return false;
}
inline bool compare2(int cur,int *pos)//wrong
{
for(int i = 0; i<n; i++)
{
*pos = i;
int ans = 0;
if(strlen(x[cur]) == strlen(z[i]))
{
for(int j = 0; j<strlen(z[i]); j++)
{
if(x[cur][j] != z[i][j])
{
ans ++;
}
}
if(ans == 1)
{
//cout<<"wrong"<<endl;
return true;
}
}
}
return false;
}

inline bool compare3(int cur,int *pos)//daoxu
{
int getc = 0;
int a[10010];
int len = strlen(x[cur]);
for(int i = 0; i<len; i++)
getc += (x[cur][i]-'a');
for(int i = 0; i<n; i++)
{
*pos = i;
int len1 = strlen(z[i]),getc1 = 0;
if(len == len1)
{
for(int j = 0; j<len1; j++)
getc1 += (z[i][j]-'a');
if(getc != getc1) continue;
else
{
int ans = 0;
for(int k = 0; k<len; k++)
{
if(x[cur][k] != z[i][k])
{
a[ans++] = k;
}
}
if(ans == 2)
{
if(x[cur][a[0]] == z[i][a[1]])
{
if(x[cur][a[1]] == z[i][a[0]])
{
//cout<<"daoxu"<<endl;
return true;
}
}
}
}
}
}
return false;
}
int main()
{
scanf("%d",&n);
for(int i = 0; i<n; i++)
scanf("%s",&z[i]);
scanf("%d",&k);
for(int i = 0; i<k; i++)
scanf("%s",&x[i]);
for(int i = 0; i<k; i++)
{
int pp = INF,pos = INF;
bool flag = false;
if(compare(i))
{
printf("%s is correct\n",x[i]);
continue;
}
if(compare1(i,&pos))
{
flag = true;
pp = min(pp,pos);
}
if(compare2(i,&pos))
{
flag = true;
pp = min(pp,pos);
}
if(compare3(i,&pos))
{
flag = true;
pp = min(pp,pos);
}
if(flag) printf("%s is a misspelling of %s\n",x[i],z[pp]);
else printf("%s is unknown\n",x[i]);
}
return 0;
}