BZOJ 4212 并非题解

ref

这个东西实际上过不了（空间炸了），但是正确性是有的。

做法：这个前缀/后缀匹配的形式容易想到在 trie 上考虑。对
原串，反串各建一个 trie，用 dfs 把有某个前缀的串的集合刻画成区间，那么问题就成了下面这个形式：在线区间查询权值在 $[x,y]$ 中的数的个数。主席树维护一下。

const int N = 2e3 + 10, L = 2e6 + 10;
bool st;
namespace trie {
    int w[L][26], sufw[L][26];
    int sz[L], dfn[L], R[L];
    int sufsz[L], sufdfn[L], sufR[L];
    int tot, dfntot, suftot, sufdfntot;
    void insert(const string& s) {
        int u = 0;
        for (auto c : s) {
            if (!w[u][c - 'a']) {
                w[u][c - 'a'] = ++tot;
            }
            u = w[u][c - 'a'];
        }
        sz[u]++;
    }
    void sufins(const string& s) {
        int u = 0;
        for (auto c : s) {
            if (!sufw[u][c - 'a']) {
                sufw[u][c - 'a'] = ++suftot;
            }
            u = sufw[u][c - 'a'];
        }
        sufsz[u]++;
    }
    void dfs(int u) {
        dfn[u] = ++dfntot;
        fir (i, 0, 26) {
            if (w[u][i]) {
                dfs(w[u][i]); sz[u] += sz[w[u][i]];
            }
        }
        R[u] = dfntot;
    }
    void sufdfs(int u) {
        sufdfn[u] = ++sufdfntot;
        fir (i, 0, 26) {
            if (sufw[u][i]) {
                sufdfs(sufw[u][i]); sufsz[u] += sufsz[sufw[u][i]];
            }
        }
        sufR[u] = sufdfntot;
    }
    int find(const string& s) {
        int u = 0;
        for (auto c : s) {
            if (w[u][c - 'a']) {
                u = w[u][c - 'a'];
            } else return -1;
        }
        return u;
    }
    int suffind(const string& s) {
        int u = 0;
        for (auto c : s) {
            if (sufw[u][c - 'a']) {
                u = sufw[u][c - 'a'];
            } else return -1;
        }
        return u;
    }
}
namespace seg {
    int w[L << 5], rt[L], ls[L << 5], rs[L << 5], tot;
    #define mid ((l + r) >> 1)
    void pushup(int u) {
        w[u] = w[ls[u]] + w[rs[u]];
    }
    void ins(int& u, int lst, int l, int r, int pos) {
        u = ++tot;
        w[u] = w[lst];
        if (l == r) {
            w[u]++; return;
        }
        if (pos <= mid) ins(ls[u], ls[lst], l, mid, pos), rs[u] = rs[lst];
        else ins(rs[u], rs[lst], mid + 1, r, pos), ls[u] = ls[lst];
        pushup(u);
    }
    int query(int u, int lst, int l, int r, int ql, int qr) {
        if (!u) return 0;
        if (ql <= l and r <= qr) return w[u] - w[lst];
        if (qr < l or r < ql) return 0;
        return query(ls[u], ls[lst], l, mid, ql, qr) + query(rs[u], rs[lst], mid + 1, r, ql, qr);
 
    }
}
using seg::rt;
bool ed;
int main() {
    #ifdef LOCAL
        assert(freopen("1.in", "r", stdin));
        assert(freopen("test.out", "w", stdout));
        assert(freopen("test.err", "w", stderr));
    #endif
    cin.tie(nullptr)->sync_with_stdio(false);
    int n;
    cin >> n;
    vla<string> s(n + 1), rs(n + 1);
    F (i, 1, n) {
        cin >> s[i];
        trie::insert(s[i]);
        rs[i] = s[i]; std::reverse(all(rs[i]));
        trie::sufins(rs[i]);
    }
    trie::sufdfs(0);
    trie::dfs(0); 
    vi a(trie::dfntot + 1);
    F (i, 1, n) {
        int u = trie::suffind(rs[i]), t = trie::dfn[trie::find(s[i])];
        a[t] = trie::sufdfn[u];
    }
    F (i, 1, trie::dfntot) {
        seg::ins(rt[i], rt[i - 1], 1, trie::dfntot, a[i]);
    }
    int Q; cin >> Q;
    int lst = 0;
    F (i, 1, Q) {
        string s1, s2; cin >> s1 >> s2;
        for (auto& c : s1) {
            c = (c - 'a' + lst) % 26 + 'a';
        }
        for (auto& c : s2) {
            c = (c - 'a' + lst) % 26 + 'a';
        }
        std::reverse(all(s2));
        int x = trie::find(s1), y = trie::suffind(s2);
        if (x == -1 or y == -1) {
            cout << "0\n"; continue;
        }
        debug(i, x, y);
        int L = trie::dfn[x], R = trie::R[x];
        int lo = trie::sufdfn[y], up = trie::sufR[y];
        debug(L, R, lo, up);
        cout << (lst = seg::query(rt[R], rt[L - 1], 1, trie::dfntot, lo, up)) << "\n";
        lst %= 26;
    }
    return 0;
}