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考虑后缀自动机。

显然我们可以求出自动机每个状态对应的子串集合在原串中出现次数，这是经典应用。

我们设其为 $f_i$，对于 $f_i \geq 2$，将 $len_i$ 算进对答案的贡献即可。复杂度线性。

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

const int N = 1e5 + 5;

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

int last = 1, tot = 1;

void extend(int c)
{
	int p = last, 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[np].fa = g[q].fa = nq;
			for (; p && g[p].son[c] == q; p = g[p].fa) g[p].son[c] = nq;
		}
	}
}
vector<int> G[N];

int t;
string s;

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

int main()
{
	ios::sync_with_stdio(0), cin.tie(nullptr), cout.tie(nullptr);
	cin >> t;
	while (t--)
	{
		for (int i = 0; i <= tot; i++) 
		{
			G[i].clear();
			g[i].len = g[i].fa = f[i] = 0;
			for (int j = 0; j < 26; j++) g[i].son[j] = 0;
		}
		cin >> s;
		last = tot = 1;
		for (char& i : s) extend(i - 'a');
		for (int i = 2; i <= tot; i++) G[g[i].fa].emplace_back(i);
		dfs(1);
		int ans = 0;
		for (int i = 1; i <= tot; i++)
		{
			if (f[i] >= 2) ans = max(ans, g[i].len);
		}
		cout << ans << "\n";
	}
	return 0;
}