给定一个长度为 nn 的字符串 SS

mm 次询问，每次询问给出 a,b,c,da,b,c,d

求 S[a,b]S[a,b]的所有子串和子串S[c,d]S[c,d]的LCP（最长公共前缀）的最大值。

n,m≤105n,m\le10^5

LOJ 2059

Solution

反着建SAM就能建出后缀树，后缀树上最两个串的lca的maxlen就是它们的lcp，具体内容可见此

反着建完SAM之后我们就是要求子串[d,c][d, c]和[b,a][b, a]中所有子串的最长公共后缀的最大值

考虑二分答案ansans，每次要判断[c−ans+1,c][c - ans + 1, c]这个子串的right集合中是否存在一个位置pp使得p∈[b+ans−1,a]p\in [b + ans - 1,a]

线段树合并求right集合即可，注意找[c−ans+1,c][c - ans + 1, c]这个子串在SAM中对应状态需要用倍增实现，具体也可见此

Code

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#include <bits/stdc++.h>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }
template <typename T> inline T read ()
{
	T sum = 0, fl = 1; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
	return sum * fl;
}

inline void proc_status ()
{
	ifstream t ("/proc/self/status");
	cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) << endl;
}

const int Maxn = 1e5 + 100;

int N, M;
char S[Maxn];

namespace SEG
{
#define mid ((l + r) >> 1)
#define ls node[root].ch[0]
#define rs node[root].ch[1]
#define lson ls, l, mid
#define rson rs, mid + 1, r

	struct info
	{
		int ch[2], cnt;
	} node[Maxn * 60];

	int node_cnt;

	inline void push_up (int root) { node[root].cnt = node[ls].cnt + node[rs].cnt; }

	inline void insert (int &root, int l, int r, int x)
	{
		++node[root = ++node_cnt].cnt;
		if (l == r) return ;
		if (x <= mid) insert (lson, x);
		else insert (rson, x);
	}

	inline int merge (int x, int y, int l, int r)
	{
		if (!x || !y) return x | y;
		int root = ++node_cnt;
		ls = merge (node[x].ch[0], node[y].ch[0], l, mid);
		rs = merge (node[x].ch[1], node[y].ch[1], mid + 1, r);
		push_up (root);
		return root;
	}

	inline int query (int root, int l, int r, int x, int y)
	{
		if (x > y) return 0;
		if (x <= l && r <= y) return node[root].cnt;

		int ans = 0;
		if (x <= mid) ans += query (lson, x, y);
		if (y > mid) ans += query (rson, x, y);
		return ans;
	}

#undef mid
#undef ls
#undef rs
#undef lson
#undef rson
}

int Root[Maxn << 1];

namespace SAM
{
	struct info
	{
		int maxlen, fa, ch[27];
	} node[Maxn << 1];

	int node_cnt = 1, last = 1;
	int State[Maxn];

	inline int new_node (int pre)
	{
		node[++node_cnt].maxlen = node[pre].maxlen + 1;
		return node_cnt;
	}

	inline int extend (int c)
	{
		int now = new_node (last), pre = last; last = now;

		for (; pre && !node[pre].ch[c]; pre = node[pre].fa) node[pre].ch[c] = now;

		if (!pre) node[now].fa = 1;
		else
		{
			int x = node[pre].ch[c];
			if (node[x].maxlen == node[pre].maxlen + 1) node[now].fa = x;
			else
			{
				int y = ++node_cnt;
				node[y] = node[x]; node[y].maxlen = node[pre].maxlen + 1;
				node[x].fa = node[now].fa = y;
				for (; node[pre].ch[c] == x; pre = node[pre].fa) node[pre].ch[c] = y;
			}
		}

		return now;
	}

	vector <int> G[Maxn << 1];
	int anc[21][Maxn << 1];

	inline void dfs (int x)
	{
		for (int i = 1; i <= 20; ++i) anc[i][x] = anc[i - 1][anc[i - 1][x]];
		for (int i = 0; i < G[x].size(); ++i)
		{
			int y = G[x][i];
			anc[0][y] = x;
			dfs (y);
			Root[x] = SEG :: merge (Root[x], Root[y], 1, node_cnt);
		}
	}

	inline void build (char *s, int n)
	{
		for (int i = 1; i <= n; ++i) State[i] = extend (s[i] - 'a');
		for (int i = 1; i <= n; ++i) SEG :: insert (Root[State[i]], 1, node_cnt, i);

		for (int i = 1; i <= node_cnt; ++i) G[node[i].fa].pb (i);
		dfs (1);
	}

	inline int Check (int ans, int a, int b, int d)
	{
		int x = State[d];
		if (node[x].maxlen < ans) return 0;
		for (int i = 20; i >= 0; --i) if (node[anc[i][x]].maxlen >= ans) x = anc[i][x];
		return SEG :: query (Root[x], 1, node_cnt, b + ans - 1, a);
	}

	inline int query (int a, int b, int c, int d)
	{
		int l = 1, r = min (a - b + 1, c - d + 1), ans = 0;
		while (l <= r)
		{
			int mid = l + r >> 1;
			if (Check (mid, a, b, c)) ans = mid, l = mid + 1;
			else r = mid - 1;
		}
		return ans;
	}
}

inline void Solve ()
{
	reverse (S + 1, S + N + 1);
	SAM :: build (S, N);

	while (M--)
	{
		int a = N - read<int>() + 1, b = N - read<int>() + 1, c = N - read<int>() + 1, d = N - read<int>() + 1;
		printf("%d\n", SAM :: query (a, b, c, d));
	}
}

inline void Input ()
{
	N = read<int>(), M = read<int>();
	scanf("%s", S + 1);
}

int main()
{

#ifndef ONLINE_JUDGE
	freopen("A.in", "r", stdin);
	freopen("A.out", "w", stdout);
#endif

	Input ();
	Solve ();

	return 0;
}