【后缀数组】【LuoguP4248】 [AHOI2013]差异

题目链接

题目描述

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

(sum_{1le i <jle n}len(T_i)+len(T_j)-2*lcp(T_i,T_j))

说明

对于 100% 的数据，保证 2⩽n⩽500000，且均为小写字母。

思路

注意到前面那个东西是个定值，所以关键在于如何求后面那个东西

由于 (lcp(sa[l],sa[r])=min_{i=l+1}^{r}H[i])

所以后面那个东西实际上就是 (H) 数组的所有子区间 (min) 之和，当然是去掉 (H[1]) 之后

这个东西可以用单调栈通过维护后缀 (min) 来求

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#define maxn 500010
#define INF 1000000000
#define ll long long
using namespace std;

int n;

char c[maxn];

int tax[maxn], rk[maxn], tp[maxn], sa[maxn], M = 200;
void rsort() {
    for (int i = 0; i <= M; ++i) tax[i] = 0;
    for (int i = 1; i <= n; ++i) ++tax[rk[i]];
    for (int i = 1; i <= M; ++i) tax[i] += tax[i - 1];
    for (int i = n; i; --i) sa[tax[rk[tp[i]]]--] = tp[i];   
}

int c1, H[maxn];
void SA(char *s) {
    for (int i = 1; i <= n; ++i) rk[i] = s[i], tp[i] = i; rsort();
    for (int k = 1; k < n; k *= 2) {
        if (c1 == n) break; M = c1; c1 = 0; 
        for (int i = n - k + 1; i <= n; ++i) tp[++c1] = i;
        for (int i = 1; i <= n; ++i) if (sa[i] > k) tp[++c1] = sa[i] - k;
        rsort(); swap(tp, rk); rk[sa[1]] = c1 = 1; 
        for (int i = 2; i <= n; ++i) {
            if (tp[sa[i - 1]] != tp[sa[i]] || tp[sa[i - 1] + k] != tp[sa[i] + k]) ++c1; 
            rk[sa[i]] = c1;
        }
    } int lcp = 0;
    for (int i = 1; i <= n; ++i) {
        if (lcp) --lcp;
        int j = sa[rk[i] - 1];
        while (s[j + lcp] == s[i + lcp]) ++lcp;
        H[rk[i]] = lcp; 
    }
}

int st[maxn], top;

ll ans, s;
int main() {
    scanf("%s", c + 1); n = strlen(c + 1); SA(c); ans = (ll) n * (n - 1) / 2 * (n + 1); 
    for (int i = 1; i < n; ++i) H[i] = H[i + 1]; H[n--] = 0;
    for (int i = 1; i <= n; ++i) {
        while (top && H[st[top]] >= H[i]) {
            s -= (ll) (st[top] - st[top - 1]) * H[st[top]];
            --top; 
        }
        st[++top] = i; s += (st[top] - st[top - 1]) * H[st[top]];
        ans -= 2 * s;
    } cout << ans << endl; 
    return 0; 
}