SPOJ LCS2 (后缀自动机)

SPOJ LCS2

Problem : 给若干个串，询问这些串的最长公共子串。
Solution :对第一个串建立后缀自动机，对于之后的每个串在后缀自动机上进行匹配。每个节点记录当前串所匹配到的长度以及所有串所匹配到的长度最小值，需要注意的是如果一个串在某个节点匹配成功，那么其fail到根的路径的所有节点都可以匹配成功，需要在最后建立出fail后缀树，进行一遍dfs进行更新。

#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 100008;

struct node
{
	int u, v, nt;
};

struct Suffix_Automaton
{
	int nt[N << 1][26], fail[N << 1], a[N << 1];
	int tot, root, last;
	int p, q, np, nq;
	node eg[N << 2];
	int lt[N << 1], sum;

	int qmax[N << 1], qmin[N << 1];

	void add(int u, int v)
	{
		eg[++sum] = (node){u, v, lt[u]}; lt[u] = sum;
		eg[++sum] = (node){v, u, lt[v]}; lt[v] = sum;
	}
	int newnode(int len)
	{
		for (int i = 0; i < 26; ++i) nt[tot][i] = -1;
		fail[tot] = -1; a[tot] = len;
		return tot++;
	}
	void clear()
	{
		last = tot = sum = 0;
		memset(lt, 0, sizeof(lt));
		root = newnode(0);
	}	
	void insert(char ch)
	{
		p = last; last = np = newnode(a[p] + 1);
		for (; ~p && nt[p][ch] == -1; p = fail[p]) nt[p][ch] = np;
		if (p == -1) fail[np] = root;
		else
		{
			q = nt[p][ch];
			if (a[q] == a[p] + 1) fail[np] = q;
			else
			{
				nq = newnode(a[p] + 1);
				for (int i = 0; i < 26; ++i) nt[nq][i] = nt[q][i];
				fail[nq] = fail[q];
				fail[q] = fail[np] = nq;
				for (; ~p && nt[p][ch] == q; p = fail[p]) nt[p][ch] = nq;
			}
		}
	}
	void build()
	{
		for (int i = 1; i < tot; ++i) add(fail[i], i);
		for (int i = 1; i < tot; ++i) qmin[i] = a[i];
	}
	void dfs(int u, int fa)
	{
		for (int i = lt[u]; i; i = eg[i].nt)
		{
			if (eg[i].v != fa)
			{
				dfs(eg[i].v, u);
				if (qmax[eg[i].v]) qmax[u] = a[u];
			}
		}
	}
	void solve(const string &s)
	{
		int p = root, cnt = 0;
		for (int i = 0; i < tot; ++i) qmax[i] = 0;
		for (int i = 0, len = s.length(); i < len; ++i)
		{
			int ch = s[i] - 'a';
			if (~nt[p][ch]) ++cnt, p = nt[p][ch];
			else
			{
				for (; ~p && nt[p][ch] == -1; p = fail[p]);
				if (p == -1) cnt = 0, p = root;
				else
				{
					cnt = a[p] + 1;
					p = nt[p][ch];
				}
			}
			qmax[p] = max(qmax[p], cnt);
		}
		dfs(root, -1);
		for (int i = 1; i < tot; ++i) qmin[i] = min(qmin[i], qmax[i]);
	}
}sam;

int main()
{
	cin.sync_with_stdio(0);
	string s;
	cin >> s;
	sam.clear();
	for (int i = 0, len = s.length(); i < len; ++i)
		sam.insert(s[i] - 'a');
	sam.build();
	while (cin >> s) sam.solve(s);
	int ans = 0;
	for (int i = 1; i < sam.tot; ++i) ans = max(ans, sam.qmin[i]);
	cout << ans << endl;
}