luogu4407 [JSOI2009]电子字典 字符串hash + hash表

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暴力枚举，然后\(hash\)表判断

复杂度\(O(26 * 20 * n)\)

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具体而言

对于操作1：暴力枚举删除

对于操作2：暴力添加，注意添加不要重复

对于操作3：暴力替换，同样的注意不要重复

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#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

#define ri register int
#define ull unsigned long long
#define rep(io, st, ed) for(ri io = st; io <= ed; io ++)
#define drep(io, ed, st) for(ri io = ed; io >= st; io --)

const int yume = 19260817;
const int sid = 4300050;

char s[50];
int n, m, L, cnp;
ull num, pre[50], wei[50];
int cap[sid], nxt[sid], val[sid];
ull fz[sid];

inline int get(ull v) {
    int k = v & 4194303;
    for(int i = cap[k]; i; i = nxt[i])
        if(fz[i] == v) return i;
    return 0;
}

inline void ins(ull v) {
    int k = v & 4194303;
    int p = get(v);
    if(!p) {
        nxt[++ cnp] = cap[k]; fz[cnp] = v;
        cap[k] = cnp; val[cnp] = 1;
    }
    else val[p] ++;
}

inline ull v(int l, int r) {
    if(l > r) return 0;
    return pre[r] - pre[l - 1] * wei[r - l + 1];
}

inline ull calc1(int p) { return v(1, p - 1) * wei[L - p] + v(p + 1, L); }
inline ull calc2(int p, char c) { return v(1, p - 1) * wei[L - p + 1] + v(p + 1, L) + c * wei[L - p]; }
inline ull calc3(int p, char c) { return v(1, p) * wei[L - p + 1] + v(p + 1, L) + c * wei[L - p]; }

int main() {
    wei[0] = 1;
    rep(i, 1, 30) wei[i] = wei[i - 1] * yume;
    cin >> n >> m;
    rep(i, 1, n) {
        scanf("%s", s + 1); L = strlen(s + 1); num = 0;
        rep(j, 1, L) num = num * yume + s[j];
        ins(num);
    }
    rep(i, 1, m) {
        scanf("%s", s + 1); L = strlen(s + 1);
        rep(j, 1, L) pre[j] = pre[j - 1] * yume + s[j];
        if(get(pre[L])) { printf("-1\n"); continue; }
        int ans = 0;
        rep(j, 1, L)
            if(s[j] != s[j - 1])
                ans += val[get(calc1(j))];
        rep(j, 1, L) rep(c, 'a', 'z')
            if(c != s[j])
                ans += val[get(calc2(j, c))];
        rep(j, 0, L) rep(c, 'a', 'z')
            if(c != s[j + 1])
                ans += val[get(calc3(j, c))];
        printf("%d\n", ans);
    }
    return 0;
}