CF666E. Forensic Examination

题意：

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[pl​..pr​]在T[l..r]中的哪个串里的出现次数最多，并输出出现次数。

如有多解输出最靠前的那一个。

解：

什么奇葩排版......

对T建广义SAM，线段树合并维护每个节点有哪些串。写法稍微注意一下。

记录下s在t上匹配的时候每个前缀走到哪个位置。然后树上倍增找到符合条件的节点，线段树上查询。

如果查到0或者根本没那么长的匹配长度，输出"L 0"

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>

const int N = 500010, M = 20000010;

struct Edge {
    int nex, v;
}edge[N * 2]; int tp;

char str[N];
int tr[N * 2][26], len[N * 2], fail[N * 2], tot = 1, n, m, L, R, e[N * 2], cnt;
std::string ss[N];
int large[M], ans[M], rt[N * 2], ls[M], rs[M], pos[N], lenth[N], fa[N * 2][20], pw[N * 2];

inline void add(int x, int y) {
    tp++;
    edge[tp].v = y;
    edge[tp].nex = e[x];
    e[x] = tp;
    return;
}

inline void Max(int &a, int b) {
    if(!b) return;
    if(large[b] > large[a] || (large[b] == large[a] && ans[a] > ans[b])) {
        a = b;
    }
    return;
}

inline void pushup(int o) {
    large[o] = std::max(large[ls[o]], large[rs[o]]);
    if(large[ls[o]] > large[rs[o]]) {
        ans[o] = ans[ls[o]];
    }
    else if(large[rs[o]] > large[ls[o]]) {
        ans[o] = ans[rs[o]];
    }
    else ans[o] = std::min(ans[ls[o]], ans[rs[o]]);
    return;
}

inline void insert(int p, int l, int r, int &o) {
    if(!o) o = ++cnt;
    if(l == r) {
        ans[o] = r;
        large[o]++;
        return;
    }
    int mid = (l + r) >> 1;
    if(p <= mid) insert(p, l, mid, ls[o]);
    else insert(p, mid + 1, r, rs[o]);
    pushup(o);
    return;
}

inline int merge(int x, int y, int l, int r) {
    if(!x || !y) return x | y;
    if(l == r) {
        int o = ++cnt;
        large[o] = large[x] + large[y];
        ans[o] = r;
        return o;
    }
    int o = ++cnt, mid = (l + r) >> 1;
    ls[o] = merge(ls[x], ls[y], l, mid);
    rs[o] = merge(rs[x], rs[y], mid + 1, r);
    pushup(o);
    return o;
}

void out(int l, int r, int x) {
    if(!x) printf("!x");
    printf("(%d [%d %d] lar=%d ans=%d) ", x, l, r, large[x], ans[x]);
    if(l == r) {
        printf("%d ", r);
        return;
    }
    int mid = (l + r) >> 1;
    if(ls[x]) out(l, mid, ls[x]);
    if(rs[x]) out(mid + 1, r, rs[x]);
    return;
}

void DFS(int x) {
    for(int i = e[x]; i; i = edge[i].nex) {
        int y = edge[i].v;
        DFS(y);
        rt[x] = merge(rt[x], rt[y], 1, m);
    }
    return;
}

inline int ask(int l, int r, int o) { /// return a Node
    if(!o) return 0;
    if(L <= l && r <= R) return o;
    int mid = (l + r) >> 1, Ans = 0;
    if(L <= mid) Max(Ans, ask(l, mid, ls[o]));
    if(mid < R) Max(Ans, ask(mid + 1, r, rs[o]));
    return Ans;
}

inline int split(int p, int f) {
    int Q = tr[p][f], nQ = ++tot;
    len[nQ] = len[p] + 1;
    fail[nQ] = fail[Q];
    fail[Q] = nQ;
    memcpy(tr[nQ], tr[Q], sizeof(tr[Q]));
    while(tr[p][f] == Q) {
        tr[p][f] = nQ;
        p = fail[p];
    }
    return nQ;
}

inline int insert(char c, int p, int id) {
    int f = c - 'a', np;
    if(tr[p][f]) {
        int Q = tr[p][f];
        if(len[Q] == len[p] + 1) {
            np = Q;
        }
        else np = split(p, f);
        insert(id, 1, m, rt[np]);
        return np;
    }
    np = ++tot;
    len[np] = len[p] + 1;
    while(p && !tr[p][f]) {
        tr[p][f] = np;
        p = fail[p];
    }
    if(!p) {
        fail[np] = 1;
    }
    else {
        int Q = tr[p][f];
        if(len[Q] == len[p] + 1) {
            fail[np] = Q;
        }
        else {
            fail[np] = split(p, f);
        }
    }
    insert(id, 1, m, rt[np]);
    return np;
}

inline int getpos(int l, int r) {
    int p = pos[r], Len = r - l + 1;
    if(lenth[r] < Len) return 0;
    int t = pw[tot];
    while(t >= 0) {
        if(len[fa[p][t]] >= Len) {
            p = fa[p][t];
        }
        t--;
    }
    return p;
}

int main() {
    scanf("%s", str + 1);
    n = strlen(str + 1);
    scanf("%d", &m);
    for(int i = 1; i <= m; i++) {
        std::cin >> ss[i];
        int k = ss[i].size(), p = 1;
        for(int j = 0; j < k; j++) {
            p = insert(ss[i][j], p, i);
        }
    }
    /// insert OVER
    for(int i = 2; i <= tot; i++) {
        add(fail[i], i);
        fa[i][0] = fail[i];
    }
    DFS(1);
    for(int i = 2; i <= tot; i++) pw[i] = pw[i >> 1] + 1;
    for(int j = 1; j <= pw[tot]; j++) {
        for(int i = 1; i <= tot; i++) {
            fa[i][j] = fa[fa[i][j - 1]][j - 1];
        }
    }
    int Len = 0, p = 1;
    for(int i = 1; i <= n; i++) {
        int ff = str[i] - 'a';
        while(p && !tr[p][ff]) {
            p = fail[p];
            Len = len[p];
        }
        if(!p) {
            p = 1;
        }
        else {
            p = tr[p][ff];
            Len++;
        }
        pos[i] = p;
        lenth[i] = Len;
    }
    /// prework OVER
    int q, l, r;
    scanf("%d", &q);
    for(int i = 1; i <= q; i++) {
        scanf("%d%d%d%d", &L, &R, &l, &r);
        p = getpos(l, r);
        if(!p) printf("%d 0
", L);
        else {
            int Ans = ask(1, m, rt[p]);
            if(!Ans) printf("%d 0
", L);
            else printf("%d %d
", ans[Ans], large[Ans]);
        }
    }
    return 0;
}

我调了一晚上差点气疯了，结果是因为tr数组第二维只开了2...这个数组脱离我的控制...