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[AHOI2013] 差异 题解

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[AHOI2013] 差异 题解

题面

题目链接

解法

使用后缀数组求出$height$数组（简写为$ht$），答案变成了

化简，得

求和符号里的用单调栈维护。

时间复杂度$O(n \log n)$。

AC代码

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/**
 * @file:           P4248.cpp
 * @author:         yaoxi-std
 * @url:            https://www.luogu.com.cn/problem/P4248
*/
// #pragma GCC optimize ("O2")
// #pragma GCC optimize ("Ofast", "inline", "-ffast-math")
// #pragma GCC target ("avx,sse2,sse3,sse4,mmx")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define resetIO(x) \
    freopen(#x ".in", "r", stdin), freopen(#x ".out", "w", stdout)
#define debug(fmt, ...) \
    fprintf(stderr, "[%s:%d] " fmt "\n", __FILE__, __LINE__, ##__VA_ARGS__)
template <class _Tp>
inline _Tp& read(_Tp &x) {
    bool sign = false;
    char ch = getchar();
    long double tmp = 1;
    for (; !isdigit(ch); ch = getchar())
        sign |= (ch == '-');
    for (x = 0; isdigit(ch); ch = getchar())
        x = x * 10 + (ch ^ 48);
    if (ch == '.')
        for (ch = getchar(); isdigit(ch); ch = getchar())
            tmp /= 10.0, x += tmp * (ch ^ 48);
    return sign ? (x = -x) : x;
}
template <class _Tp>
inline void write(_Tp x) {
    if (x < 0)
        putchar('-'), x = -x;
    if (x > 9)
        write(x / 10);
    putchar((x % 10) ^ 48);
}
#define fi first
#define se second
using pii = pair<int, int>;
const int MAXN = 5e5 + 10;
const int INF = 0x3f3f3f3f3f3f3f3f;
struct suffix_array {
    int n, sa[MAXN], rk[MAXN], tp[MAXN], ht[MAXN];
    void radix_sort(int m) {
        static int buk[MAXN];
        for (int i = 0; i <= m; ++i)
            buk[i] = 0;
        for (int i = 1; i <= n; ++i)
            buk[rk[i]]++;
        for (int i = 1; i <= m; ++i)
            buk[i] += buk[i - 1];
        for (int i = n; i >= 1; --i)
            sa[buk[rk[tp[i]]]--] = tp[i];
    }
    void init(char *s, int n) {
        this->n = n;
        int m = 100;
        for (int i = 1; i <= n; ++i)
            rk[i] = s[i] - 'a' + 1, tp[i] = i;
        radix_sort(m);
        for (int p = 0, w = 1; p < n; m = p, w <<= 1) {
            p = 0;
            for (int i = 1; i <= w; ++i)
                tp[++p] = n - w + i;
            for (int i = 1; i <= n; ++i)
                if (sa[i] > w)
                    tp[++p] = sa[i] - w;
            radix_sort(m);
            copy(rk + 1, rk + n + 1, tp + 1);
            rk[sa[1]] = p = 1;
            for (int i = 2; i <= n; ++i) {
                if (tp[sa[i - 1]] == tp[sa[i]] && tp[sa[i - 1] + w] == tp[sa[i] + w])
                    rk[sa[i]] = p;
                else
                    rk[sa[i]] = ++p;
            }
        }
        for (int i = 1, k = 0; i <= n; ++i) {
            if (k)
                k--;
            while (s[i + k] == s[sa[rk[i] - 1] + k])
                k++;
            ht[i] = k;
        }
    }
};
int n, top;
pii sta[MAXN];
char s[MAXN];
suffix_array sa;
signed main() {
    scanf("%s", s + 1);
    n = strlen(s + 1);
    sa.init(s, n);
    int ans = 0, sum = 0;
    sta[top = 1] = {0, 1};
    for (int i = 2; i <= n; ++i) {
        int p = sa.sa[i];
        while (top && sta[top].fi >= sa.ht[p])
            sum -= sta[top].fi * (sta[top].se - sta[top - 1].se), --top;
        sta[++top] = {sa.ht[p], i}, sum += sta[top].fi * (sta[top].se - sta[top - 1].se);
        ans += sum;
    }
    ans = (n * (n + 1) * (n - 1)) / 2 - ans * 2;
    write(ans), putchar('\n');
    return 0;
}