题意

题目链接

Sol

说一个后缀自动机+线段树的无脑做法

首先建出SAM，然后对parent树进行dp，维护最大次大值，最小次小值

显然一个串能更新答案的区间是\([len_{fa_{x}} + 1, len_x]\)，方案数就相当于是从\(siz_x\)里面选两个，也就是\(\frac{siz_x (siz_x - 1)}{2}\)

直接拿线段树维护一下，标记永久化一下炒鸡好写~

#include<bits/stdc++.h>
#define int long long 
#define LL long long 
using namespace std;
const int MAXN = 1e6 + 10;
const LL INF = 2e18 + 10;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, a[MAXN];
char s[MAXN];
int root = 1, tot = 1, las = 1, ch[MAXN][26], fa[MAXN], len[MAXN], rev[MAXN];
LL mx[MAXN], mx2[MAXN], ans[MAXN], ans1[MAXN], mn[MAXN], mn2[MAXN], tmp[MAXN], siz[MAXN];
vector<int> v[MAXN];
void insert(int x, int id) {
    int now = ++tot, pre = las; las = now; siz[now] = 1; len[now] = len[pre] + 1; mx[now] = a[id]; mx2[now] = -INF; mn[now] = a[id]; mn2[now] = INF; rev[id] = now;
    for(; pre && !ch[pre][x]; pre = fa[pre]) ch[pre][x] = now;
    if(!pre) {fa[now] = root; return ;}
    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]));
        fa[now] = fa[q] = nq;
        for(; pre && ch[pre][x] == q; pre = fa[pre]) ch[pre][x] = nq;
    }
}
void BuildDAG() {
    for(int i = 1; i <= tot; i++) assert(fa[i] != i), v[fa[i]].push_back(i);
}

int rt, Node, ls[MAXN], rs[MAXN], ad[MAXN], si[MAXN];
LL sum[MAXN], tag[MAXN];
void Build(int &k, int l, int r) {
    if(!k) k = ++Node, tag[k] = -INF, si[k] = r - l + 1;
    if(l == r) return ;
    int mid = l + r >> 1;
    Build(ls[k], l, mid);
    Build(rs[k], mid + 1, r);
}
void IntMax(int k, int l, int r, int ql, int qr, LL v) {
    if(ql <= l && r <= qr) {chmax(tag[k], v); return ;  }
    int mid = l + r >> 1;
    if(ql <= mid) IntMax(ls[k], l, mid, ql, qr, v);
    if(qr  > mid) IntMax(rs[k], mid + 1, r, ql, qr, v);
}
void IntAdd(int k, int l, int r, int ql, int qr, LL v) {
    if(ql <= l && r <= qr) {sum[k] += v; return ;}
    int mid = l + r >> 1;
    if(ql <= mid) IntAdd(ls[k], l, mid, ql, qr, v);
    if(qr  > mid) IntAdd(rs[k], mid + 1, r, ql, qr, v);
}
LL QueryNum(int k, int l, int r, int pos) {
    if(!k) return 0;
    LL now = sum[k];
    if(l == r || !k) return now;
    int mid = l + r >> 1;
    if(pos <= mid) now += QueryNum(ls[k], l, mid, pos);
    else now += QueryNum(rs[k], mid + 1, r, pos);
    return now;
}
LL QueryMax(int k, int l, int r, int pos) {
    if(!k) return -INF;
    LL now = tag[k];
    if(l == r || !k) return now;
    int mid = l + r >> 1;
    if(pos <= mid) chmax(now, QueryMax(ls[k], l, mid, pos));
    else chmax(now, QueryMax(rs[k], mid + 1, r, pos));
    return now;
}
void dfs(int x) {
    for(auto &to : v[x]) {
        dfs(to);
        siz[x] += siz[to];
        if(mx2[to] > mx[x]) chmax(mx2[x], mx[x]), mx[x] = mx2[to];
        else chmax(mx2[x], mx2[to]);
        if(mx[to] > mx[x]) chmax(mx2[x], mx[x]), mx[x] = mx[to];
        else chmax(mx2[x], mx[to]);
        
        if(mn2[to] < mn[x]) chmin(mn2[x], mn[x]), mn[x] = mn2[to];
        else chmin(mn2[x], mn2[to]);
        if(mn[to] < mn[x]) chmin(mn2[x], mn[x]), mn[x] = mn[to];
        else chmin(mn2[x], mn[to]); 
    }
    if(siz[x] > 1 && x != root) {
        IntMax(rt, 1, N, len[fa[x]] + 1, len[x], mx[x] * mx2[x]);
        IntMax(rt, 1, N, len[fa[x]] + 1, len[x], mn