POJ 3518 (后缀自动机)

POJ 3518 Boring

Problem : 给一个串S，询问串S有多个子串出现至少两次且位置不重叠。

Solution : 对S串建立后缀自动机，再建立后缀树，dfs一遍统计处每个结点的子树中最长节点max和最短节点min。枚举一遍后缀自动机的节点，那么对于其对应后缀的长度要求为小于等于max - min。

#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1000008;
const int INF = 2000000008;
struct edge
{
	int u, v, nt;
};
struct suffix_automanon
{
	int nt[N][26], fail[N], a[N], qmin[N], qmax[N];
	int tot, last, root;
	int lt[N], sum;
	int p, q, np, nq;
	edge eg[N << 1];
	void add(int u, int v)
	{
		eg[++sum] = (edge){u, v, lt[u]}; lt[u] = sum;
	}
	int newnode(int len)
	{
		for (int i = 0; i < 26; ++i) nt[tot][i] = -1;
		fail[tot] = -1; a[tot] = len; qmax[tot] = 0; qmin[tot] = INF;
		lt[tot] = 0;
		return tot++;
	}
	void clear()
	{
		tot = 0;
		root = last = newnode(0);
	}
	void insert(int ch)
	{
		p = last; np = last = newnode(a[p] + 1); qmin[np] = qmax[np] = a[np];
		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[p] + 1 == a[q]) 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 dfs(int u)
	{
		for (int i = lt[u]; i; i = eg[i].nt)
		{
	//		cout << u << " " << eg[i].v << endl;
			int v = eg[i].v;
			dfs(v);
			qmax[u] = max(qmax[u], qmax[v]);
			qmin[u] = min(qmin[u], qmin[v]);
		}
	}
	void solve()
	{
		long long ans = 0;
		for (int i = 1; i < tot; ++i) add(fail[i], i);
		dfs(root);
	//	for (int i = 1; i < tot; ++i) cout << qmin[i] << " " << qmax[i] << endl;
		for (int i = 1; i < tot; ++i)
		{
			int len = qmax[i] - qmin[i];
			if (len > a[fail[i]]) ans += min(a[i], len) - a[fail[i]];
		}
		cout << ans << endl;
	}
}sam;
int main()
{
	string s;
	while (cin >> s)
	{
		if (s == "#") break;
		sam.clear();
		for (int i = 0, len = s.length(); i < len; ++i)
			sam.insert(s[i] - 'a');
		sam.solve();
	}
}