CF666E Forensic Examination

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给你一个串$S$以及一个字符串数组$T[1..m]$，$q$次询问，每次问$S$的子串$S[p_l..p_r]$在$T[l..r]$中的哪个串里的出现次数最多，并输出出现次数。

考虑求查询子串$[q_l,q_r]$在$T_i$出现的次数。和求最长公共子串一样，匹配不了往上跳$link$。对于找到的往上倍增跳$link$,使得公共长度维持在$\geq pr-pl+1$。（尽可能让$endpos$大），然后求当前$endpos$里出现次数。

多个子串可以建立广义后缀自动机，对于每个串建立最后权值线段树，维护$endpos$里出现次数最多的$i$，线段树合并。

  • 预处理$S[1..i]$，最多匹配的长度和位置。
  • 由于$maxlen[link[]]$满足单调性，可以倍增一下。

复杂度$O(|S|+Q\times (\log|T|+\log m)$

代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
#define ls tree[pos].l
#define rs tree[pos].r
const int N = 1e6 + 10;
const int mod = 1e9 + 7;

const int SN = 1e6 + 100;

struct seg
{
    int l, r, maxx;
    int siz;
} tree[SN * 20];
int scnt;
void pushup(int pos)
{
    tree[pos].maxx = (tree[ls].siz >= tree[rs].siz ? tree[ls].maxx : tree[rs].maxx);
    tree[pos].siz = max(tree[ls].siz, tree[rs].siz);
}
void modify(int &pos, int l, int r, int w)
{
    if (!pos)
        pos = ++scnt;
    if (l == r)
    {
        tree[pos].siz++;
        tree[pos].maxx = l;
        return;
    }

    int mid = (l + r) >> 1;
    if (w <= mid)
        modify(ls, l, mid, w);
    else
        modify(rs, mid + 1, r, w);
    pushup(pos);
}
pii query(int pos, int ql, int qr, int l, int r)
{
    if (ql <= l && r <= qr)
        return mk(tree[pos].maxx, tree[pos].siz);
    int mid = (l + r) >> 1;
    pii res = mk(0, 0);
    if (ql <= mid)
    {
        pii now = query(ls, ql, qr, l, mid);
        if (now.second >= res.second)
            res = now;
    }
    if (qr > mid)
    {
        pii now = query(rs, ql, qr, mid + 1, r);
        if (now.second > res.second)
            res = now;
    }
    return res;
}

int merge(int u, int v, int l, int r)
{
    if (!u || !v)
        return u + v;
    int pos = ++scnt;
    if (l == r)
    {
        tree[pos].siz = tree[u].siz + tree[v].siz;
        tree[pos].maxx = l;
        return pos;
    }
    int mid = (l + r) >> 1;
    ls = merge(tree[u].l, tree[v].l, l, mid);
    rs = merge(tree[u].r, tree[v].r, mid + 1, r);
    pushup(pos);
    return pos;
}

int rt[SN], m;
struct SAM
{
    int trans[SN][26];
    int maxlen[SN], link[SN], pre, scnt;
    SAM() { pre = scnt = 1; };
    void init()
    {
        for (int j = 1; j <= scnt; j++)
        {
            link[j] = 0;
            maxlen[j] = 0;
            //siz[j] = 0;
            memset(trans[j], 0, sizeof(trans[j]));
        }
        scnt = pre = 1;
    };
    void insert(int id, int p)
    {
        int cur = ++scnt;
        maxlen[cur] = maxlen[pre] + 1;
        modify(rt[cur], 1, m, p);
        int u;

        for (u = pre; u && !trans[u][id]; u = link[u])
            trans[u][id] = cur;
        pre = cur;
        if (!u)
            link[cur] = 1;
        else
        {
            int x = trans[u][id];
            if (maxlen[x] == maxlen[u] + 1)
                link[cur] = x;
            else
            {
                int nc = ++scnt;
                link[nc] = link[x];
                maxlen[nc] = maxlen[u] + 1;
                memcpy(trans[nc], trans[x], sizeof(trans[x]));

                link[cur] = link[x] = nc;
                for (; u && trans[u][id] == x; u = link[u])
                    trans[u][id] = nc;
            }
        }
    }
} sam;
int f[N][30];
struct edge
{
    int to, nxt;
} e[SN];
int ecnt, head[SN];
void addadge(int u, int v)
{
    e[++ecnt].to = v;
    e[ecnt].nxt = head[u];
    head[u] = ecnt;
}
void dfs(int x)
{
    for (int i = 1; i < 20; i++)
        f[x][i] = f[f[x][i - 1]][i - 1];
    for (int i = head[x]; i; i = e[i].nxt)
    {
        int to = e[i].to;
        f[to][0] = x;
        dfs(to);
        rt[x] = merge(rt[x], rt[to], 1, m);
    }
}

char s[N], t[N];
int ed[N], lp[N];
void prepare()
{
    int u = 1, l = 0;
    int slen = strlen(s + 1);
    for (int i = 1; i <= slen; i++)
    {
        int id = s[i] - 'a';
        while (u && !sam.trans[u][id])
        {
            u = sam.link[u];
            l = sam.maxlen[u];
        }
        if (sam.trans[u][id])
            u = sam.trans[u][id], l++;
        else
            u = 1, l = 0;
        ed[i] = u, lp[i] = l;
    }
    // for (int i = 1; i <= slen; i++)
    //     cout << lp[i] << endl;
}
pii solve(int l, int r, int pl, int pr)
{
    if (pr - pl + 1 > lp[pr])
        return mk(l, 0);
    int u = ed[pr];

    for (int i = 19; i >= 0; i--)
        if (f[u][i] && sam.maxlen[f[u][i]] >= pr - pl + 1)
            u = f[u][i];

    pii ans = query(rt[u], l, r, 1, m);
    if (ans.second == 0)
        return mk(l, 0);
    else
        return ans;
}
int main()
{
    scanf("%s", s + 1);

    scanf("%d", &m);
    for (int i = 1; i <= m; i++)
    {
        scanf("%s", t + 1);
        sam.pre = 1;
        int le = strlen(t + 1);
        for (int j = 1; j <= le; j++)
            sam.insert(t[j] - 'a', i);
    }
    for (int j = 2; j <= sam.scnt; j++)
        addadge(sam.link[j], j);
    dfs(1);
    prepare();
    int q;
    scanf("%d", &q);
    while (q--)
    {
        int l, r, pl, pr;
        scanf("%d %d %d %d", &l, &r, &pl, &pr);
        pii ans = solve(l, r, pl, pr);
        printf("%d %d\n", ans.first, ans.second);
    }
}