Fake News (hard)

什么板子能评 $2300$。

建完 SAM，对于每一个点，其表示字符串个数为 $len_u - len_{fa_u}$，出现次数可以通过 $\operatorname{link}$ 边，即后缀链接线性求出，也是套路了。

然后算贡献就没了，复杂度线性。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
using namespace std;

const int N = 2e5 + 5;

int tot, last;
int f[N];
long long ans = 0;
int q;
vector<int> G[N];

struct Node
{
	int son[26], len, fa;
}g[N];

void extend(int c)
{
	int p = last;
	int np = last = ++tot;
	g[np].len = g[p].len + 1;
	f[np] = 1;
	for (; p && g[p].son[c] == 0; p = g[p].fa) g[p].son[c] = np;
	if (!p) g[np].fa = 1;
	else
	{
		int q = g[p].son[c];
		if (g[q].len == g[p].len + 1) g[np].fa = q;
		else
		{
			int nq = ++tot;
			g[nq] = g[q];
			g[nq].len = g[p].len + 1;
			g[q].fa = g[np].fa = nq;
			for (; p && g[p].son[c] == q; p = g[p].fa) g[p].son[c] = nq;
		}
	}
}

void dfs1(int u)
{
	for (int j : G[u])
	{
		dfs1(j);
		f[u] += f[j];
	}
}

int main()
{
	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
	cin >> q;
	while (q--)
	{
		for (int i = 1; i <= tot; i++) 
		{
			f[i] = 0;
			g[i].len = g[i].fa = 0;
			for (int j = 0; j < 26; j++) g[i].son[j] = 0;
			G[i].clear();
		}
		tot = last = 1;
		string s;
		cin >> s;
		for (auto i : s)
		{
			extend(i - 'a');
		}
		for (int i = 2; i <= tot; i++) G[g[i].fa].emplace_back(i);
		ans = 0;
		dfs1(1);
		for (int i = 2; i <= tot; i++)
		{
			ans += f[i] * 1LL * f[i] * (g[i].len * 1LL - g[g[i].fa].len);
		}
		cout << ans << "\n";
	}
	return 0;
}