SPOJ Lexicographical Substring Search 后缀自动机

给你一个字符串，然后询问它第k小的factor，坑的地方在于spoj实在是太慢了，要加各种常数优化，字符集如果不压缩一下必t。。

#pragma warning(disable:4996)

#include<cstring>

#include<string>

#include<iostream>

#include<cmath>

#include<vector>

#include<algorithm>

#define maxn 90050

using namespace std;



struct State{

	State *suf, *go[26];

	int val, cnt;

	char transch;

	State() :suf(0), val(0){

		memset(go, 0, sizeof(go));

	}

}*root, *last;



State statePool[maxn * 2], *cur;



void init()

{

	cur = statePool;

	root = last = cur++;

}



void extend(int w)

{

	State *p = last, *np = cur++;

	np->val = p->val + 1;

	np->cnt = 1;

	while (p&&!p->go[w]) p->go[w] = np, p = p->suf;

	if (!p) np->suf = root;

	else{

		State *q = p->go[w];

		if (p->val + 1 == q->val){

			np->suf = q;

		}

		else{

			State *nq = cur++;

			memcpy(nq->go, q->go, sizeof q->go);

			nq->val = p->val + 1;

			nq->cnt = 1;

			nq->suf = q->suf;

			q->suf = nq;

			np->suf = nq;

			while (p&&p->go[w] == q){

				p->go[w] = nq, p = p->suf;

			}

		}

	}

	last = np;

}



char str[maxn];

char ans[maxn];

int atop;

int n;

int q;



int bcnt[maxn];

State *b[maxn * 2];



int main()

{

	scanf("%s", str);

	n = strlen(str);

	init();

	for (int i = 0; i < n; i++){

		extend(str[i] - 'a');

	}

	int tot = cur - statePool;

	for (int i = 0; i < tot; i++) bcnt[statePool[i].val]++;

	for (int i = 1; i <= n; i++) bcnt[i] += bcnt[i - 1];

	for (int i = 0; i < tot; i++) b[--bcnt[statePool[i].val]] = statePool + i;

	for (int i = tot - 1; i >= 0; i--){

		int kth = 0;

		State *p = b[i];

		for (int j = 0; j < 26; j++){

			if (p->go[j]){

				p->cnt += p->go[j]->cnt;

				p->go[kth++] = p->go[j];

				p->go[j]->transch = 'a' + j;

			}

		}

		p->go[kth] = NULL;

	}

	scanf("%d", &q);

	int k;

	while (q--)

	{

		scanf("%d", &k);

		State *p = root; atop = 0;

		while (k){

			for (int x = 0; p->go[x]; x++){

				if (k > p->go[x]->cnt) k -= p->go[x]->cnt;

				else {

					k -= 1;

					p = p->go[x];

					ans[atop++] = p->transch;

					break;

				}

			}

		}

		ans[atop] = '\0';

		puts(ans);

	}

	return 0;

}