[AHOI 2013]差异

Description

题库链接

给定一个长度为 (n) 的字符串 (S) ，令 (T_i) 表示它从第 (i) 个字符开始的后缀。求

[sum_{1leqslant i<jleqslant n} ext{len}(T_i)+ ext{len}(T_j)-2 imes ext{lcp}(T_i,T_j)]

其中， ( ext{len}(a)) 表示字符串 (a) 的长度， ( ext{lcp}(a,b)) 表示字符串 (a) 和字符串 (b) 的最长公共前缀。

(2leqslant nleqslant 500000) ，且均为小写字母。

Solution

注意到原式可化为

[egin{aligned}&(n-1)sum_{i=1}^n ext{len}(T_i)-2sum_{1leqslant i<jleqslant n} ext{lcp}(T_{sa_i},T_{sa_j})\=&frac{(n-1)n(n+1)}{2}-2sum_{1leqslant i<jleqslant n} ext{lcp}(T_{sa_i},T_{sa_j})end{aligned}]

记 ( ext{LCP}(i,j)= ext{lcp}(T_{sa_i},T_{sa_j})) ，由于

[ ext{LCP}(i,j)=min_{i<kleq j} ext{LCP}(k-1,k)]

那么就可以单调栈预处理出两个数组 (l_i,r_i) 表示左边（右边）第一个大于（大于等于） (height_i) 的位置。注意，由于不能重复计算，等于只能一边取。

然后直接算贡献就好了。

Code

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 500000+5;

char ch[N];
int n, m, x[N<<1], y[N<<1], c[N], sa[N], rk[N], height[N], l[N], r[N], s[N], top;

void get_sa() {
    for (int i = 1; i <= n; i++) c[x[i] = ch[i]]++;
    for (int i = 2; i <= m; i++) c[i] += c[i-1];
    for (int i = n; i >= 1; i--) sa[c[x[i]]--] = i;
    for (int k = 1; k <= n; k <<= 1) {
        int num = 0;
        for (int i = n-k+1; i <= n; i++) y[++num] = i;
        for (int i = 1; i <= n; i++) if (sa[i] > k) y[++num] = sa[i]-k;
        for (int i = 1; i <= m; i++) c[i] = 0;
        for (int i = 1; i <= n; i++) c[x[i]]++;
        for (int i = 2; i <= m; i++) c[i] += c[i-1];
        for (int i = n; i >= 1; i--) sa[c[x[y[i]]]--] = y[i];
        swap(x, y); x[sa[1]] = num = 1;
        for (int i = 2; i <= n; i++)
            x[sa[i]] = (y[sa[i]] == y[sa[i-1]] && y[sa[i]+k] == y[sa[i-1]+k]) ? num : ++num;
        if ((m = num) == n) break;
    }
}
void get_height() {
    for (int i = 1; i <= n; i++) rk[sa[i]] = i;
    for (int i = 1, k = 0; i <= n; i++) {
        if (rk[i] == 1) continue;
        if (k) --k; int j = sa[rk[i]-1];
        while (j+k <= n && i+k <= n && ch[i+k] == ch[j+k]) ++k;
        height[rk[i]] = k;
    }
}
void work() {
    scanf("%s", ch+1), n = strlen(ch+1), m = 'z';
    get_sa(); get_height(); ll ans = 1ll*(1+n)*n*(n-1)/2;
    s[++top] = n+1;
    for (int i = n; i >= 2; i--) {
        while (top != 1 && height[i] <= height[s[top]]) --top;
        r[i] = s[top]; s[++top] = i;
    }
    s[top = 1] = 1;
    for (int i = 2; i <= n; i++) {
        while (top != 1 && height[i] < height[s[top]]) --top;
        l[i] = s[top]; s[++top] = i;
    }
    for (int i = 2; i <= n; i++) ans -= 2ll*(i-l[i])*(r[i]-i)*height[i];
    printf("%lld
", ans);
}
int main() {work(); return 0; }