[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 + 1), len = strlen(s + 1), ++ cnt;
    for(R i = 1; i <= len; i ++) hs[cnt] = hs[cnt] * base + s[i];
}

void pre()
{
    scanf("%d%d", &n, &m);
    for(R i = 1; i <= n; i ++) build();
    sort(hs + 1, hs + n + 1);
    qw[0] = 1;
    for(R i = 1; i <= 20; i ++) qw[i] = qw[i - 1] * base;
}

bool half(us LL x)
{
    int l = 1, r = n, mid;
    while(l < r)
    {
        mid = (l + r) >> 1;
        if(hs[mid] == x) 
        {
            if(id[mid] == dfn) return false;
            id[mid] = dfn; return true;
        }
        else if(hs[mid] < x) l = mid + 1;
        else r = mid - 1;
    }
    if(hs[l] != x) return false;
    if(id[l] == dfn) return false;
    id[l] = dfn; return true;
}

bool check()
{
    scanf("%s", s + 1), len = strlen(s + 1), ans = 0, ++ dfn;
    us LL x = 0;
    memset(rs, 0, 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 = 1; i <= len; i ++) x = x * base + s[i];
    if(half(x))
    {
        printf("-1
");
        return true;
    }
    for(R i = 1; i <= len; i ++) ls[i] = ls[i - 1] * base + s[i];
    for(R i = 1; i <= len; i ++) rs[i] = ls[len] - ls[i - 1] * qw[len - i + 1];
    //for(R i = 1; i <= len; i ++) printf("%lld ", rs[i]);
    //printf("
");
    return false;
}

void get()
{
    us LL x;
    for(R i = 0; 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 = 0; j < 26; j ++)//枚举修改成哪个
            {
                x = ls[i - 1] * qw[len - i + 1] + 1LL * (j + 'a') * qw[len - i] + rs[i + 1];
                if(half(x)) ans ++;                
            }
            x = ls[i - 1] * qw[len - i] + rs[i + 1];
            if(half(x)) ans ++;
        }
        for(R j = 0; j < 26; j ++)//枚举在后面插入哪个
        {
            x = ls[i] * qw[len - i + 1] + 1LL * (j + 'a') * qw[len - i] + rs[i + 1];
            if(half(x)) ans ++;
        }
    }
    printf("%d
", ans);
}

void work()
{
    for(R i = 1; i <= m; i ++) 
    {
        if(check()) continue;
        get();
    }
}

int main()
{
//    freopen("in.in", "r", stdin);
    pre();
    work();
//    fclose(stdin);
    return 0;
}