「AHOI 2013」差异 「SA」「单调栈」

首先明确题目要我们求什么。看到后面的 \(LCP(T_i, T_j)\) 很容易用后缀数组将其转化成 \(\min_{rk[i] < k \leq rk[j]}{height[k]}\)。\((若rk[i] < rk[j])\)

考虑计算每个位置的h作为min出现的次数。很明显这个东西可以用单调栈一步求出来。那么就转为计算

\(\sum_{p = l} ^ {i} \sum_{p = i} ^ {r} (n - sa[i - 1] + 1) + (n - sa[i] + 1)\)。

然后大家只要有脑子就可以发现上面那个式子加起来是\(\sum_{i = 1} ^ {n} i * (n - 1)\)，因为任何一个长度为\(i\)的后缀都会被计算\(n-1\)次。

那么这个题就做完了，注意单调栈的l,r求法并不相同，因为要考虑去掉重复。

#include <bits/stdc++.h>

#pragma GCC optimize(2)

#define rep(i, l, r) for (int i = (l); i <= (r); ++i)
#define per(i, r, l) for (int i = (r); i >= (l); --i)
using namespace std;

typedef long long ll;
typedef pair <int, int> pii;
typedef vector <int> vi;

int gi() {
    int f = 1, x = 0; char ch = getchar();
    while (ch < '0' || ch > '9') {if (ch == '-') f = -1; ch = getchar();}
    while (ch >= '0' && ch <= '9') {x = x * 10 + ch - '0'; ch = getchar();}
    return f * x;
}

const int N = 500010;
int sa[N], rk[N], px[N], cnt[N], rk_[N], id[N], h[N], n;
int st[N], tp, l[N], r[N];
char s[N];
ll pref[N];
bool cmp(int x, int y, int w) {
    return rk_[x] == rk_[y] && rk_[x + w] == rk_[y + w];
}
void get_SA() {
    int i, j, m = 300, p, w;
    for (i = 1; i <= n; ++i) ++cnt[rk[i] = s[i]];
    for (i = 1; i <= m; ++i) cnt[i] += cnt[i - 1];
    for (i = n; i >= 1; --i) sa[cnt[rk[i]]--] = i;
    for (w = 1; w < n; w <<= 1, m = p) {
        for (p = 0, i = n; i > n - w; --i) id[++p] = i;
        for (i = 1; i <= n; ++i) if (sa[i] > w) id[++p] = sa[i] - w;
        memset(cnt, 0, sizeof(cnt));
        for (i = 1; i <= n; ++i) ++cnt[px[i] = rk[id[i]]];
        for (i = 1; i <= m; ++i) cnt[i] += cnt[i - 1];
        for (i = n; i >= 1; --i) sa[cnt[px[i]]--] = id[i];
        memcpy(rk_, rk, sizeof(rk));
        for (p = 0, i = 1; i <= n; ++i)
            rk[sa[i]] = cmp(sa[i - 1], sa[i], w) ? p : ++p;
    }
    for (i = 1, j = 0; i <= n; ++i) {
        if (rk[i] == 1) continue;
        int pos = sa[rk[i] - 1];
        while (s[i + j] == s[pos + j]) ++j;
        h[rk[i]] = j;
        if (j) --j;
    }
//  cout << "height: ";
//  for (i = 1; i <= n; ++i) cout << h[i] << " ";
//  cout << '\n';
}
ll SUM(ll l, ll r) {
    return pref[r] - pref[l - 1];
}
int main() {
    scanf ("%s", s + 1);
    n = strlen(s + 1);
    get_SA();
    rep (i, 1, n) l[i] = 0, r[i] = n + 1;
    for (int i = 1; i <= n; ++i) {
        while (tp && h[st[tp]] >= h[i]) --tp;
        if (tp) l[i] = st[tp];
        st[++tp] = i;
    }
    tp = 0;
    for (int i = n; i >= 1; --i) {
        while (tp && h[st[tp]] > h[i]) --tp;
        if (tp) r[i] = st[tp];
        st[++tp] = i;
    }
    for (int i = 1; i <= n; ++i) pref[i] = pref[i - 1] + sa[i];
    ll ans = 0, ANS = 0;
    for (int i = 1; i <= n; ++i) ans += 1LL * i * (n - 1);
    for (int i = 1; i <= n; ++i) {
        int seg_l = l[i] + 1, seg_r = r[i] - 1;
        ll ways = (i - seg_l + 1) * 1ll * (seg_r - i + 1);
        ANS += ways;
        ans += -2LL * h[i] * ways;
    }
    cout << ans << '\n';
  return 0;
}

原文地址：https://www.cnblogs.com/LiM-817/p/12302143.html