后缀数组笔记。

Height 数组应用 3。

子串等价于后缀的前缀，考虑容斥，用总数减去重复数量。
则答案为 $\frac{n(n+1)}{2} - \sum\limits_{i=2}^{n} height_i$，因为 $height$ 的定义是 $\text{LCP}$ 长度。

#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int T, n;
char s[N];

int sa[N], rk[N], x[N], y[N];
int cnt[N];
void init_SA() {
    int v = 128;
    for (int i = 0; i <= v; i++) cnt[i] = 0;
    for (int i = 1; i <= n; i++) ++cnt[x[i] = s[i]];
    for (int i = 1; i <= v; i++) cnt[i] += cnt[i - 1];
    for (int i = n; i >= 1; i--) sa[ cnt[x[i]]-- ] = i;

    for (int len = 1; ; len <<= 1) {
        int tot = 0;
        for (int i = n - len + 1; i <= n; i++) y[++tot] = i;
        for (int i = 1; i <= n; i++)
            if (sa[i] > len) y[++tot] = sa[i] - len;

        for (int i = 0; i <= v; i++) cnt[i] = 0;
        for (int i = 1; i <= n; i++) ++cnt[x[i]];
        for (int i = 1; i <= v; i++) cnt[i] += cnt[i - 1];
        for (int i = n; i >= 1; i--) sa[ cnt[x[y[i]]]-- ] = y[i], y[i] = 0;
        swap(x, y), tot = 0;
        for (int i = 1; i <= n; i++) {
            if (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + len] == y[sa[i - 1] + len]) x[sa[i]] = tot;
            else x[sa[i]] = ++tot;
        }
        v = tot;
        if (v == n) break;
    }
}

int ht[N];
void init_height() {
    for (int i = 1; i <= n; i++) rk[sa[i]] = i;
    for (int i = 1, j = 0, now = 0; i <= n; i++) {
        if (rk[i] == 1) { ht[rk[i]] = 0; continue; }
        j = sa[rk[i] - 1];
        if (now) now--;
        while (i + now <= n && j + now <= n && s[i + now] == s[j + now]) now++;
        ht[rk[i]] = now;
    }
}

long long ans = 0;
void solve() {
    scanf("%s", s + 1), n = strlen(s + 1);
    init_SA(), init_height();
    ans = n * 1ll * (n + 1) / 2;
    for (int i = 2; i <= n; i++) ans -= ht[i];
    printf("%lld\n", ans);
}

int main() {
    scanf("%d", &T);
    while (T--) solve();
    return 0;
}