[JSOI2009]电子字典 hash

题面：洛谷

题解：

　　做法。。。。非常暴力。

　　因为要求的编辑距离最多只有1，所以我们直接枚举对那个位置（字符）进行操作，进行什么样的操作，加入/修改/删除哪个字符，然后暴力枚举hash判断即可，

 #include<bits/stdc++.h>
 using namespace std;
 #define R register int
 #define AC 25
 #define ac 10100
 #define base 26
 #define LL long long
 #define us unsigned

 int n, m, len, ans, cnt, dfn;
 int id[ac];
 us LL hs[ac], qw[AC], ls[AC], rs[AC];
 char s[AC];

 void build()
 {
     scanf("%s", s + ), len = strlen(s + ), ++ cnt;
     for(R i = ; i <= len; i ++) hs[cnt] = hs[cnt] * base + s[i];
 }

 void pre()
 {
     scanf("%d%d", &n, &m);
     for(R i = ; i <= n; i ++) build();
     sort(hs + , hs + n + );
     qw[] = ;
     for(R i = ; i <= ; i ++) qw[i] = qw[i - ] * base;
 }

 bool half(us LL x)
 {
     int l = , r = n, mid;
     while(l < r)
     {
         mid = (l + r) >> ;
         if(hs[mid] == x)
         {
             if(id[mid] == dfn) return false;
             id[mid] = dfn; return true;
         }
         else if(hs[mid] < x) l = mid + ;
         else r = mid - ;
     }
     if(hs[l] != x) return false;
     if(id[l] == dfn) return false;
     id[l] = dfn; return true;
 }

 bool check()
 {
     scanf("%s", s + ), len = strlen(s + ), ans = , ++ dfn;
     us LL x = ;
     memset(rs, , sizeof(rs));//why?????大概是因为下面判断的时候可能用到r[len + 1]吧，而r[len + 1]应该要=0的
     for(R i = ; i <= len; i ++) x = x * base + s[i];
     if(half(x))
     {
         printf("-1\n");
         return true;
     }
     for(R i = ; i <= len; i ++) ls[i] = ls[i - ] * base + s[i];
     for(R i = ; i <= len; i ++) rs[i] = ls[len] - ls[i - ] * qw[len - i + ];
     //for(R i = 1; i <= len; i ++) printf("%lld ", rs[i]);
     //printf("\n");
     return false;
 }

 void get()
 {
     us LL x;
     for(R i = ; i <= len; i ++)//枚举对哪一位进行操作，
     {//如果是插入的话就是在这个位置的后面插入
         if(i)//只有i > 0才可以删除和修改
         {
             for(R j = ; j < ; j ++)//枚举修改成哪个
             {
                 x = ls[i - ] * qw[len - i + ] + 1LL * (j + 'a') * qw[len - i] + rs[i + ];
                 if(half(x)) ans ++;
             }
             x = ls[i - ] * qw[len - i] + rs[i + ];
             if(half(x)) ans ++;
         }
         for(R j = ; j < ; j ++)//枚举在后面插入哪个
         {
             x = ls[i] * qw[len - i + ] + 1LL * (j + 'a') * qw[len - i] + rs[i + ];
             if(half(x)) ans ++;
         }
     }
     printf("%d\n", ans);
 }

 void work()
 {
     for(R i = ; i <= m; i ++)
     {
         if(check()) continue;
         get();
     }
 }

 int main()
 {
 //    freopen("in.in", "r", stdin);
     pre();
     work();
 //    fclose(stdin);
     return ;
 }