题意
两个长度为$r$的子串相等称为$r$相似，两个$r$相似的权值等于子串开头位置权值乘积，给定字符串和每个位置权值，求$r$相似子串数量和最大权值乘积
对反串建立后缀自动机得到后缀树，后缀树上两个状态的lca的长度len就是原串的两个子串的lcp，在树上进行dp，parent树上每个状态代表长度为$maxlen_s-maxlen_{pa_s}$长度的一段子串，对于相似子串数量，$r$相似子串的数量$ans_r+=Right_u\times Right_v，maxlen_u==r$，对于最大权值乘积，维护每个状态的最大值最小值，每次用最大值最大值的乘积和最小值乘最小值的乘积更新答案，最后将答案累计
时间复杂度$O(n)$
代码
#include <bits/stdc++.h>
#define inf (LL)1000000000000000001
using namespace std;
typedef long long LL;
const int N = 1001000;
int ch[N][30], pa[N], maxlen[N], sz, last, Right[N], mark[N], val[N];
inline void init_sam() {
memset(ch, 0, sizeof(ch));
last = sz = 1;
}
inline void extend(int c, int x) {
int p = last, np = ++sz; last = np; maxlen[np] = maxlen[p] + 1; mark[np] = 1; val[np] = x;
while(p && !ch[p][c]) ch[p][c] = np, p = pa[p];
if(!p) {pa[np] = 1; return;}
int q = ch[p][c];
if(maxlen[q] == maxlen[p] + 1) {
pa[np] = q;
}else {
int nq = ++sz;
memcpy(ch[nq], ch[q], sizeof(ch[q]));
pa[nq] = pa[q]; maxlen[nq] = maxlen[p] + 1; pa[q] = pa[np] = nq;
while(ch[p][c] == q) ch[p][c] = nq, p = pa[p];
}
}
int cnt, head[N], nxt[N], to[N];
inline void init_edge() {cnt = 0; memset(head, -1, sizeof(head));}
inline void add(int u, int v) {to[cnt] = v; nxt[cnt] = head[u]; head[u] = cnt++;}
int n, a[N];
LL ans1[N], ans2[N], Max[N], Min[N];
char str[N];
void dfs(int u) {
Right[u] = 0; Max[u] = -inf; Min[u] = inf;
if(mark[u]) Right[u] = 1, Max[u] = Min[u] = val[u];
for(int i = head[u]; ~i; i = nxt[i]) {
int v = to[i]; dfs(v);
if(Max[u] != -inf && Max[v] != -inf && Min[u] != inf && Min[v] != inf) {
ans2[maxlen[u]] = max(ans2[maxlen[u]], max(Max[u] * Max[v], Min[u] * Min[v]));
}
Max[u] = max(Max[u], Max[v]); Min[u] = min(Min[u], Min[v]);
ans1[maxlen[u]] += 1LL * Right[u] * Right[v];
Right[u] += Right[v];
}
}
int main() {
init_sam();
init_edge();
scanf("%d", &n);
scanf("%s", str + 1);
for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
for(int i = n; i >= 1; --i) extend(str[i] - 'a', a[i]);
for(int i = 2; i <= sz; ++i) add(pa[i], i);
for(int i = 0; i <= n; ++i) ans2[i] = -inf;
dfs(1);
for(int i = n - 1; i >= 0; --i) ans1[i] += ans1[i + 1], ans2[i] = max(ans2[i], ans2[i + 1]);
for(int i = 0; i < n; ++i) {
if(ans1[i])printf("%lld %lld\n", ans1[i], ans2[i]);
else printf("0 0\n");
}
return 0;
}