POJ 1035 Spell checker(字符串）

题目链接

考虑各种情况，然后注意敲的稳一点，忘考虑一种情况,2Y.

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <queue>
  5 #include <string>
  6 #include <map>
  7 using namespace std;
  8 char word[10001][21];
  9 char s[21];
 10 int len[10001];
 11 int Abs(int x)
 12 {
 13     if(x < 0)
 14     return -x;
 15     else
 16     return x;
 17 }
 18 int main()
 19 {
 20     int i,j,n,k1,k,L,z;
 21     map<string,int>mp;
 22     for(i = 1;;i ++)
 23     {
 24         scanf("%s",word[i]);
 25         len[i] = strlen(word[i]);
 26         if(word[i][0] == '#')
 27         break;
 28         mp[word[i]] = 1;
 29     }
 30     n = i-1;
 31     for(i = 1;;i ++)
 32     {
 33         scanf("%s",s);
 34         if(s[0] == '#') break;
 35         L = strlen(s);
 36         if(mp[s])
 37         {
 38             printf("%s is correct\n",s);
 39             continue;
 40         }
 41         printf("%s:",s);
 42         for(j = 1;j <= n;j ++)
 43         {
 44             if((len[j]-L) > 1)
 45             continue;
 46             if(len[j] == L)
 47             {
 48                 z = 0;
 49                 for(k = 0;k < L;k ++)
 50                 {
 51                     if(s[k] != word[j][k])
 52                     z ++;
 53                     if(z >= 2)
 54                     break;
 55                 }
 56                 if(z == 1)
 57                 printf(" %s",word[j]);
 58             }
 59             else if(len[j] > L)
 60             {
 61                 z = 0;
 62                 for(k = 0,k1 = 0;k < L;k ++,k1 ++)
 63                 {
 64                     if(s[k] != word[j][k1])
 65                     {
 66                         if(s[k] == word[j][k1+1])
 67                         k1 ++;
 68                         else
 69                         break;
 70                         z ++;
 71                     }
 72                 }
 73                 if(k == L&&k1 == len[j]&&z == 1)
 74                 printf(" %s",word[j]);
 75                 else if(k == L&&k1 == len[j]-1&&z == 0)
 76                 printf(" %s",word[j]);
 77             }
 78             else
 79             {
 80                 z = 0;
 81                 for(k = 0,k1 = 0;k < L;k ++,k1 ++)
 82                 {
 83                     if(s[k] != word[j][k1])
 84                     {
 85                         if(s[k+1] == word[j][k1])
 86                         k ++;
 87                         else
 88                         break;
 89                         z ++;
 90                     }
 91                 }
 92                 if(k == L&&k1 == len[j]&&z == 1)
 93                 printf(" %s",word[j]);
 94                 else if(k == L+1&&k1 == len[j]+1&&z == 1)//随便试了一组，这里就查出错了。。aaa aaaa
 95                 printf(" %s",word[j]);
 96             }
 97         }
 98         printf("\n");
 99     }
100     return 0;
101 }

 

 

posted @ 2012-12-13 11:31  Naix_x  阅读(135)  评论(0)  编辑  收藏  举报