我们定义2个字符串的相似度等于两个串的相同前缀的长度。例如 “abc” 同 “abd” 的相似度为2，“aaa” 同 “aaab” 的相似度为3。
给出一个字符串S，计算S同他所有后缀的相似度之和。例如：S = “ababaa”，所有后缀为：

ababaa 6
babaa 0
abaa 3
baa 0
aa 1
a 1

S同所有后缀的相似度的和 = 6 + 0 + 3 + 0 + 1 + 1 = 11

输入
输入一个字符串S(1 <= L <= 1000000)，L为字符串S的长度，且S由a-z的小写字母组成。
输出
输出S同所有后缀的相似度的和。
输入样例
ababaa
输出样例
11

后缀数组模板题，倍增算法O（nlogn）复杂度的过不去，需要DC3算法O(n)复杂度才能卡过去（正解是exKMP算法）但是我不会

倍增算法

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 5;
char s[maxn];
int height[maxn], sa[maxn], tx[maxn], rak[maxn];
struct node {
	int x, y, id;
}a[maxn], b[maxn];
int n, m, p;
void rsort() {
	for (int i = 1; i <= m; i++) {
		tx[i] = 0;
	}
	for (int i = 1; i <= n; i++) {
		tx[a[i].y]++;
	}
	for (int i = 1; i <= m; i++) {
		tx[i] += tx[i - 1];
	}
	for (int i = 1; i <= n; i++) {
		b[tx[a[i].y]--] = a[i];
	}
	for (int i = 1; i <= m; i++) {
		tx[i] = 0;
	}
	for (int i = 1; i <= n; i++) {
		tx[b[i].x]++;
	}
	for (int i = 1; i <= m; i++) {
		tx[i] += tx[i - 1];
	}
	for (int i = n; i >= 1; i--) {
		a[tx[b[i].x]--] = b[i];
	}
}


void ssort() {
	rsort();
	p = 0;
	for (int i = 1; i <= n; i++) {
		if (a[i].x != a[i - 1].x || a[i].y != a[i - 1].y) {
			++p;
		}
		rak[a[i].id] = p;
	}
	for (int i = 1; i <= n; i++) {
		a[i].x = rak[i];
		a[i].id = sa[rak[i]] = i;
		a[i].y = 0;
	}
	m = p;
}

void solve() {
	m = 127;
	for (int i = 1; i <= n; i++) {
		a[i].x = a[i].y = s[i];
		a[i].id = i;
	}
	ssort();
	for (int j = 1; j <= n; j <<= 1) {
		for (int i = 1; i + j <= n; i++) {
			a[i].y = a[i + j].x;
		}
		ssort();
		if (p == n) {
			break;
		}
	}
}

void get_Height() {
	int k = 0;
	for (int i = 1; i <= n; i++) {
		if (k) {
			k--;
		}
		int j = sa[rak[i] - 1];
		while (i + k <= n && j + k <= n && s[i + k] == s[j + k]) {
			k++;
		}
		height[rak[i]] = k;
	}
}


int main() {
	scanf("%s", s + 1);
	n = strlen(s + 1);
	solve(); get_Height();
	long long ans = 0;
	int mm = 1e9;
	for (int i = rak[1] + 1; i <= n; i++) {
		mm = min(mm, height[i]);
		ans += mm;
	}
	mm = 1e9;
	for (int i = rak[1]; i >= 1; i--) {
		mm = min(mm, height[i]);
		ans += mm;
	}
	cout << ans + n << endl;
	return 0;
}

DC3算法

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
#define rank Rank
#define F(x) ((x)/3+((x)%3==1?0:tb))
#define G(x) ((x)<tb?(x)*3+1:((x)-tb)*3+2)
using namespace std;
const int maxn=1e6*3+5;
int n;
char s[maxn];
int sa[maxn],rank[maxn],height[maxn];
int r[maxn],wa[maxn],wb[maxn],wv[maxn],wws[maxn];
void sort(int *r,int *a,int *b,int n,int m)
{
    int i;
    for(i=0; i<n; i++) wv[i]=r[a[i]];
    for(i=0; i<m; i++) wws[i]=0;
    for(i=0; i<n; i++) wws[wv[i]]++;
    for(i=1; i<m; i++) wws[i]+=wws[i-1];
    for(i=n-1; i>=0; i--) b[--wws[wv[i]]]=a[i];
    return;
}
int c0(int *r,int a,int b)
{
    return r[a]==r[b]&&r[a+1]==r[b+1]&&r[a+2]==r[b+2];
}
int c12(int k,int *r,int a,int b)
{
    if(k==2) return r[a]<r[b]||r[a]==r[b]&&c12(1,r,a+1,b+1);
    else return r[a]<r[b]||r[a]==r[b]&&wv[a+1]<wv[b+1];
}
void get_sa_dc3(int *r,int *sa,int n,int m)
{
    int i,j,*rn=r+n,*san=sa+n,ta=0,tb=(n+1)/3,tbc=0,p;
    r[n]=r[n+1]=0;
    for(i=0; i<n; i++) if(i%3!=0) wa[tbc++]=i;
    sort(r+2,wa,wb,tbc,m);
    sort(r+1,wb,wa,tbc,m);
    sort(r,wa,wb,tbc,m);
    for(p=1,rn[F(wb[0])]=0,i=1; i<tbc; i++) rn[F(wb[i])]=c0(r,wb[i-1],wb[i])?p-1:p++;
    if(p<tbc) get_sa_dc3(rn,san,tbc,p);
    else for(i=0; i<tbc; i++) san[rn[i]]=i;
    for(i=0; i<tbc; i++) if(san[i]<tb) wb[ta++]=san[i]*3;
    if(n%3==1) wb[ta++]=n-1;
    sort(r,wb,wa,ta,m);
    for(i=0; i<tbc; i++) wv[wb[i]=G(san[i])]=i;
    for(i=0,j=0,p=0; i<ta && j<tbc; p++) sa[p]=c12(wb[j]%3,r,wa[i],wb[j])?wa[i++]:wb[j++];
    for(; i<ta; p++) sa[p]=wa[i++];
    for(; j<tbc; p++) sa[p]=wb[j++];
    return;
}
void get_height(int *r, int *sa, int n)
{
    int i, j, k = 0;
    for (i = 1; i <= n; ++i) rank[sa[i]] = i;
    for (i = 0; i < n; height[rank[i++]] = k)
        for (k ? k-- : 0, j = sa[rank[i] - 1]; r[i + k] == r[j + k]; ++k);
    return;
}

int main()
{
	scanf("%s", s);
    int Max=1227,ans=1;
    n=strlen(s);
    for(int i=0;i<n;i++) r[i]=s[i];
    r[n]=0;
    get_sa_dc3(r,sa,n+1,Max+1);
    get_height(r,sa,n);
    long long res = 0;
    int mm = 1e9;
    for (int i = rank[0] + 1; i <= 3 * n; i++) {
    	mm = min(mm, height[i]);
    	res += mm;
	}
	mm = 1e9;
	for (int i = rank[0]; i >= 0; i--) {
		mm = min(mm, height[i]);
		res += mm;
	}
	printf("%lld\n", res + n);
    return 0;
}