Description

(我并不想告诉你题目名字是什么鬼)

有一个长度为n的仅包含小写字母的字符串S,下标范围为[1,n].

现在有若干组询问,对于每一个询问,我们给出若干个后缀(以其在S中出现的起始位置来表示),求这些后缀两两之间的LCP(LongestCommonPrefix)的长度之和.一对后缀之间的LCP长度仅统计一遍

Input

第一行两个正整数n,m,分别表示S的长度以及询问的次数.

接下来一行有一个字符串S.

接下来有m组询问,对于每一组询问,均按照以下格式在一行内给出:

首先是一个整数t,表示共有多少个后缀.接下来t个整数分别表示t个后缀在字符串S中的出现位置

Output

对于每一组询问,输出一行一个整数,表示该组询问的答案.由于答案可能很大,仅需要输出这个答案对于23333333333333333(一个巨大的质数)取模的余数.

Sample Input

7 3

popoqqq

1 4

2 3 5

4 1 2 5 6

Sample Output

0

0

2

### Hint
样例解释:
对于询问一,只有一个后缀”oqqq”,因此答案为0.
对于询问二,有两个后缀”poqqq”以及”qqq”,两个后缀之间的LCP为0,因此答案为0.
对于询问三,有四个后缀”popoqqq”,”opoqqq”,”qqq”,”qq”,其中只有”qqq”,”qq”两个后缀之间的LCP不为0,且长度为2,因此答案为2.
对于100%的测试数据,有\(S<=5*10^5\),且\(\sum t<=3*10^6\).
特别注意:由于另一世界线的某些参数发生了变化,对于一组询问,即使一个后缀出现了多次,也仅算一次.

首先一个很显然的思路就是直接把这个数组按照rank排序，然后我们发现对于每个点，前面的点的贡献从前往后是单调不减的，然后就可以直接维护单调栈了

挺水的题

#include<bits/stdc++.h>using namespace std;typedef pair<int, int> pi;
typedef long long ll;
const int N = 5e5 + 10;
const int M = 3e6 + 10;
const int LOG = 20;
const ll Mod = 23333333333333333;struct Suffix_Array {int s[N], n, m;int c[N], x[N], y[N];int height[N], sa[N], rank[N];int st[N][LOG], Log[N];ll sum[N]; void init(int len, char *c) {n = len, m = 0;for (int i = 1; i <= len; i++) {s[i] = c[i];m = max(m, s[i]);}}void radix_sort() {for (int i = 1; i <= m; i++) c[i] = 0;for (int i = 1; i <= n; i++) c[x[y[i]]]++;for (int i = 1; i <= m; i++) c[i] += c[i - 1];for (int i = n; i >= 1; i--) sa[c[x[y[i]]]--] = y[i];}void buildsa() {for (int i = 1; i <= n; i++) x[i] = s[i], y[i] = i;radix_sort();int now;for (int k = 1; k <= n; k <<= 1) {now = 0;for (int i = n - k + 1; i <= n; i++) y[++now] = i;for (int i = 1; i <= n; i++) if (sa[i] > k) y[++now] = sa[i] - k;radix_sort();y[sa[1]] = now = 1;for (int i = 2; i <= n; i++) y[sa[i]] = (x[sa[i]] == x[sa[i - 1]] && x[sa[i] + k] == x[sa[i - 1] + k]) ? now : ++now;swap(x, y);if (now == n) break;m = now;}}void buildrank() {for (int i = 1; i <= n; i++) rank[sa[i]] = i;}void buildsum() {for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + n - sa[i] + 1 - height[i];}void buildheight() {for (int i = 1; i <= n; i++) if (rank[i] != 1) {int k = max(height[rank[i - 1]] - 1, 0); // 里面是 rank[i - 1] for (; s[i + k] == s[sa[rank[i] - 1] + k]; k++);height[rank[i]] = k; // height 里面是 rank }}void buildst() {Log[1] = 0;for (int i = 2; i < N; i++) Log[i] = Log[i >> 1] + 1;for (int i = 1; i <= n; i++) st[i][0] = height[i];for (int j = 1; j < LOG; j++) {for (int i = 1; i + (1 << (j - 1)) <= n; i++) {st[i][j] = min(st[i][j - 1], st[i + (1 << (j - 1))][j - 1]);}}}int queryst(int l, int r) {if (l == r) return n - sa[l] + 1;if (l > r) swap(l, r);++l; //***int k = Log[r - l + 1];return min(st[l][k], st[r - (1 << k) + 1][k]);}int querylcp(int la, int lb) {return queryst(rank[la], rank[lb]);}void build(int len, char *c) {init(len, c);buildsa();buildrank();buildheight();buildsum();buildst();}
} Sa;int n, q, m;
char c[N];
ll ans, sum;
pi p[M];struct Node {int num, pos;ll val;
};
stack<Node> Q;ll add(ll a, ll b) {return (a += b) >= Mod ? a - Mod : a;
}int main() {
#ifdef dream_makerfreopen("input.txt", "r", stdin);
#endifscanf("%d %d", &n, &q);scanf("%s", c + 1);Sa.build(n, c); while (q--) {scanf("%d", &m);for (int i = 1; i <= m; i++) {scanf("%d", &p[i].second);p[i].first = Sa.rank[p[i].second];}sort(p + 1, p + m + 1);m = unique(p + 1, p + m + 1) - p - 1;ans = sum = 0;while (Q.size()) Q.pop();Q.push((Node) {1, p[1].second, 0});for (int i = 2; i <= m; i++) {int curnum = 1, len = Sa.querylcp(Q.top().pos, p[i].second);while (Q.size() && Q.top().val >= len) {curnum += Q.top().num;sum -= Q.top().val * Q.top().num;Q.pop();}Q.push((Node) {curnum, p[i].second, len});sum += len * curnum; ans = add(ans, sum);}printf("%lld\n", ans);}return 0;
}