poj 1035 Spell checker

 

题意:首先给定一些字符串(字典),然后再输入一个字符串,然后在字典里面查找看是否有,有则输出

没有则需要你纠正,纠正的原则是

替换一个字符

插入一个字符

删除一个字符

 

 

View Code

  1 /*

  2     poj1035

  3     题意:给定一些字典里面的字符串,以#表示结束,然后再给定一个字符串str

  4                 若在字典里面能够找得到则输出:cout<<str<<" is correct"<<endl;

  5                 若没有,则按如下规则:

  6                     删除一个字母在字典里是否有

  7                     添加一个字母在字典里是否有

  8                     替换一个字母在字典里是否有.                

  9 

 10 */

 11 #include <iostream>

 12 #include <string>

 13 using namespace std;

 14 string dir[10005];

 15 int dirlen[10005];

 16 string str;

 17 int len;    //str.length()

 18 int fix[10005], f;

 19 void Change(int k)

 20 {

 21     int pos = 0;

 22     for (int i=0; i<len; i++)

 23     {

 24         if (str[i] != dir[k][i])    //不相同则替换掉.

 25             pos ++;

 26         if (pos == 2)

 27             break;

 28     }

 29     if (pos != 2)

 30         fix[f++] = k;

 31 }

 32 void Delete(int k)    // len -1 == dirlen[k]

 33 {

 34     int pos = 0;

 35     for (int i=0,j=0; i<len && j<dirlen[k]; i++, j++ )

 36     {

 37         if (str[i] != dir[k][j])

 38         {

 39             j--;

 40             pos ++;

 41             if (pos == 2)

 42                 break;

 43         }

 44     }

 45     if (pos != 2)

 46         fix[f++] = k;

 47 }

 48 void Add(int k)    //len + 1 == dirlen[k]

 49 {

 50     int pos = 0;

 51     for (int i=0,j=0; j<dirlen[k] && i<len; i++, j++)

 52     {

 53         if (str[i] != dir[k][j])

 54         {

 55             i--;

 56             pos ++;

 57             if (pos == 2)

 58                 break;

 59         }

 60     }

 61     if (pos != 2)

 62         fix[f++] = k;

 63 }

 64 int main()

 65 {

 66     int n=0,i;

 67     while (cin>>dir[n] && dir[n][0] != '#')

 68     {

 69         dirlen[n] = dir[n].length();

 70         n++;

 71     }

 72     while (cin>>str && str[0] != '#')

 73     {

 74         len = str.length();

 75         f = 0;

 76         bool flg = false;

 77         for (i=0; i<n; i++)

 78         {

 79             if (len == dirlen[i])

 80             {

 81                 if(str.compare(dir[i]) == 0)

 82                 {

 83                     flg = true;

 84                     break;

 85                 }

 86                 else 

 87                     Change(i);

 88             }

 89             else if (len - 1 == dirlen[i])

 90                 Delete(i);

 91             else if (len + 1 == dirlen[i])

 92                 Add(i);

 93         }

 94         if (flg)

 95             cout<<str<<" is correct"<<endl;

 96         else

 97         {

 98             cout<<str<<":";

 99             for (i=0; i<f; i++)

100                 cout<<" "<<dir[fix[i]];

101             cout<<endl;

102         }

103     }

104     return 0;

105 }