BZOJ3413: 匹配(后缀自动机 线段树合并)

题意

sol

神仙题orz

后缀自动机 + 线段树合并。。。

首先可以转化一下模型(想不到qwq)：问题可以转化为统计\(b\)中每个前缀在\(a\)中出现的次数。(画一画就出来了)

然后直接对\(a\)串建sam，线段树合并维护一下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»´î³öïöµäî»öã 
    }
    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
*/