BZOJ3413: 匹配(尾碼自動機 線段樹合併)

題意

題目鏈接

Sol

神仙題Orz

尾碼自動機 + 線段樹合併。。。

首先可以轉化一下模型(想不到qwq)：問題可以轉化為統計\(B\)中每個首碼在\(A\)中出現的次數。(畫一畫就出來了)

然後直接對\(A\)串建SAM，線段樹合併維護一下siz就行了

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 4e5 + 10, SS = 1e7 + 10;
int N, M;
char S[MAXN], T[MAXN];
int fa[MAXN], len[MAXN], ch[MAXN][11], root = 1, las = 1, tot = 1;
vector<int> par[MAXN];
int insert(int x) {
    int now = ++tot, pre = las; las = now; len[now] = len[pre] + 1;
    for(; pre && !ch[pre][x]; pre = fa[pre]) ch[pre][x] = now;
    if(!pre) fa[now] = root;
    else {
        int q = ch[pre][x];
        if(len[pre] + 1 == len[q]) fa[now] = q;
        else {
            int nq = ++tot; fa[nq] = fa[q]; len[nq] = len[pre] + 1;
            memcpy(ch[nq], ch[q], sizeof(ch[q]));
            for(; pre && ch[pre][x] == q; pre = fa[pre]) ch[pre][x] = nq;
            fa[q] = fa[now] = nq;
        }
    }
    return las;
}
void Build() {
    for(int i = 1; i <= tot; i++) par[fa[i]].push_back(i);
}
int rt[SS], ls[SS], rs[SS], sum[SS], cnt;
void update(int k) {
    sum[k] = sum[ls[k]] + sum[rs[k]];
}
void Modify(int &k, int l, int r, int p, int v) {
    if(!k) k = ++cnt;
    if(l == r) {sum[k]++; return ;}
    int mid = l + r >> 1;
    if(p <= mid) Modify(ls[k], l, mid, p, v);
    else Modify(rs[k], mid + 1, r, p, v);
    update(k);
}
int Merge(int x, int y) {
    if(!x || !y) return x ^ y;
    int nw = ++cnt;
    if(!ls[x] && !rs[x]) {sum[nw] = sum[x] + sum[y]; return nw;}
    ls[nw] = Merge(ls[x], ls[y]);
    rs[nw] = Merge(rs[x], rs[y]);
    update(nw);
    return nw;
}
int Get(int k, int l, int r) {
    if(!k) return N;
    if(l == r) return l;
    int mid = l + r >> 1;
    if(sum[ls[k]]) return Get(ls[k], l, mid);
    else return Get(rs[k], mid + 1, r);
}
int Query(int k, int l, int r, int ql, int qr) {
    if(!k || (l > r) || (ql > qr)) return 0;
    if(ql <= l && r <= qr) 
        return sum[k];
    int mid = l + r >> 1;
    if(ql > mid) return Query(rs[k], mid + 1, r, ql, qr);
    else if(qr <= mid) return Query(ls[k], l, mid, ql, qr);
    else return Query(ls[k], l, mid, ql, qr) + Query(rs[k], mid + 1, r, ql, qr);
}
void dfs(int x) {
    for(auto &to : par[x]) {
        dfs(to);
        rt[x] = Merge(rt[x], rt[to]);
    }
}
void solve() {
    int n = strlen(T + 1), now = root, flag = 0, Lim = 0;
    for(int i = 1; i <= n; i++) {
        int nxt = T[i] - '0';
        if(!ch[now][nxt]) {flag = 1; break;}
        now = ch[now][nxt];
        if(i == n) 
            Lim = Get(rt[now], 1, N) - n;//µÚÒ»´Î³öÏÖµÄλÖà 
    }
    int ans = 0;
    if(flag) ans = N;
    else ans = Lim + n;
    now = root;
    for(int i = 1; i <= n; i++) {
        int nxt = T[i] - '0';
        if(!ch[now][nxt]) break;
        now = ch[now][nxt];
        if(flag) ans += Query(rt[now], 1, N, 1, N);
        else ans += Query(rt[now], 1, N, 1, Lim + i - 1);
    }
    cout << ans << '\n';
}
int main() {
    //freopen("1.in", "r", stdin); freopen("b.out", "w", stdout);
    cin >> N;
    scanf("%s", S + 1);
    for(int i = 1; i <= N; i++) 
        Modify(rt[insert(S[i] - '0')], 1, N, i, 1);
    Build();
    dfs(root);
    cin >> M;
    for(int i = 1; i <= M; i++) {
        scanf("%s", T + 1);
        solve();
    }
    return 0;
}
/*
7
1090901
4
0901
87650
109
090
*/