SPOJ Lexicographical Substring Search 后缀自动机

最新推荐文章于 2020-02-01 18:18:56 发布

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最新推荐文章于 2020-02-01 18:18:56 发布

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原文链接：http://www.cnblogs.com/chanme/p/3551952.html

本文探讨了一个在处理字符串操作时遇到性能瓶颈的问题，特别是如何通过优化字符串处理和压缩字符集来提高SPJ（Single Pass Join）算法的效率。通过引入状态机和常数优化策略，有效地减少了计算时间，确保了算法在处理大规模数据时的性能。

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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;
}

转载于:https://www.cnblogs.com/chanme/p/3551952.html