反复折腾了POJ上的后缀数组

折腾了最长回文串,和模式串的匹配问题。


发现程序慢如狗!


POJ 3461 TLE


#include <cstdio>
#include <cstring>


const int max_strlen = 1200000 + 10;
char pattern[max_strlen], text[max_strlen];
int sa[max_strlen], tub[max_strlen], wa[max_strlen], wb[max_strlen], wv[max_strlen];
int R[max_strlen], height[max_strlen], rank[max_strlen];
int totlen, SA[max_strlen];
int plen, tlen;
bool cmp(int *r, int a, int b, int l)
{return r[a] == r[b] && r[a + l] == r[b + l];}


void da(int *r, int *sa, int n, int m)
{
	int i, j, p, *x = wa, *y = wb, *t;
	for (i = 0; i != m; ++ i)	tub[i] = 0;
	for (i = 0; i != n; ++ i)	++ tub[x[i] = r[i]];
	for (i = 1; i != m; ++ i)	tub[i] += tub[i - 1];
	for (i = n - 1; i >= 0; -- i)	sa[-- tub[x[i]]] = i;
	for (j = 1, p =1; p != n; m = p, j *= 2)
	{
		for (p = 0, i = n - j; i!= n; ++ i)	y[p ++] = i;	
		for (i = 0; i != n; ++ i)	if (sa[i] >= j)	y[p ++] = sa[i] - j;
		for (i = 0; i != n; ++ i)	wv[i] = x[y[i]];
		for (i = 0; i != m; ++ i)	tub[i] = 0;
		for (i = 0; i != n; ++ i)	++ tub[wv[i]];
		for (i = 1; i != m; ++ i)	tub[i] += tub[i - 1];
		for (i = n - 1; i >= 0; -- i)	sa[-- tub[wv[i]]] = y[i];
		for (t = x, x =y, y = t, p = 1, x[sa[0]] = 0, i = 1; i != n; ++ i)
			x[sa[i]] = cmp(y, sa[i], sa[i - 1], j) ? p - 1 : p ++;
	}
}

void calheight(int *r, int *sa, int n)
{
	int i, j, k = 0;
	for (i = 1; i <= n; ++ i)	rank[sa[i]] = i;
	for (i = 0; i != n; height[rank[i ++ ]] = k)
		for (j = sa[rank[i] - 1], k ? k -- : 0; r[i + k] == r[j + k]; ++ k);
}

void HZSZ()
{
	totlen = 0;
	for (int i = 0; i != plen; ++ i)	R[totlen ++] = pattern[i];
	R[totlen ++] = 4; //串分割符号
	for (int i = 0; i != tlen; ++ i)	R[totlen ++] = text[i];
	R[totlen] = 3; //结束符号
	da(R, SA, totlen + 1, 175);
	calheight(R, SA, totlen);
}

void doit()
{
	int pos = rank[0];
	int ans = 0;
	for (int i = pos; i >= 1; -- i)
	{
		if (height[i] < plen)	break;	
		++ ans;
	}
	for (int i = pos + 1; i <= totlen; ++ i)
	{
		if (height[i] < plen)	break;
		++ ans;	
	}
	printf("%d\n", ans);
}


int main()
{
	int n;
	scanf("%d\n", &n);
	while (n --)
	{
		gets(pattern);
		gets(text);	
		plen = strlen(pattern);
		tlen = strlen(text);
		HZSZ(); //cal出SA和height数组
		doit();
	}
	return 0;
}



POJ 3974,直接无法承受后缀数组所需要的内存空间,直接爆内存!

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;


const int max_strlen = 4000000  + 10;
char text[max_strlen];
int tub[max_strlen], wa[max_strlen], wb[max_strlen], wv[max_strlen];
int R[max_strlen], height[max_strlen], rank[max_strlen];
int totlen, SA[max_strlen];
int textlen;
bool cmp(int *r, int a, int b, int l)
{return r[a] == r[b] && r[a + l] == r[b + l];}


void da(int *r, int *sa, int n, int m)
{
	int i, j, p, *x = wa, *y = wb, *t;
	for (i = 0; i != m; ++ i)	tub[i] = 0;
	for (i = 0; i != n; ++ i)	++ tub[x[i] = r[i]];
	for (i = 1; i != m; ++ i)	tub[i] += tub[i - 1];
	for (i = n - 1; i >= 0; -- i)	sa[-- tub[x[i]]] = i;
	for (j = 1, p =1; p != n; m = p, j *= 2)
	{
		for (p = 0, i = n - j; i!= n; ++ i)	y[p ++] = i;	
		for (i = 0; i != n; ++ i)	if (sa[i] >= j)	y[p ++] = sa[i] - j;
		for (i = 0; i != n; ++ i)	wv[i] = x[y[i]];
		for (i = 0; i != m; ++ i)	tub[i] = 0;
		for (i = 0; i != n; ++ i)	++ tub[wv[i]];
		for (i = 1; i != m; ++ i)	tub[i] += tub[i - 1];
		for (i = n - 1; i >= 0; -- i)	sa[-- tub[wv[i]]] = y[i];
		for (t = x, x =y, y = t, p = 1, x[sa[0]] = 0, i = 1; i != n; ++ i)
			x[sa[i]] = cmp(y, sa[i], sa[i - 1], j) ? p - 1 : p ++;
	}
}

void calheight(int *r, int *sa, int n)
{
	int i, j, k = 0;
	for (i = 1; i <= n; ++ i)	rank[sa[i]] = i;
	for (i = 0; i != n; height[rank[i ++ ]] = k)
		for (j = sa[rank[i] - 1], k ? k -- : 0; r[i + k] == r[j + k]; ++ k);
}

void HZSZ()
{
	totlen = 0;
	R[totlen ++] = 5;//开头符号
	R[totlen ++] = 6;

	for (int i = 0; i != textlen; ++ i)	
	{
		R[totlen ++] = text[i];
		R[totlen ++] = 6; //字符中断符
	}
	R[totlen ++] = 4; //串分割符号
	R[totlen ++] = 6;
	for (int i = textlen - 1; i >=0 ; -- i)
	{
		R[totlen ++] = text[i];
		R[totlen ++] = 6;
	}
	R[totlen] = 3; //结束符号
	da(R, SA, totlen + 1, 175);
	calheight(R, SA, totlen);
}

struct node
{
	node *ls, *rs;	
	int key, L, R;
	node(int LL, int RR, int KEY, node *LS, node *RS)
	{
		ls = LS;
		rs = RS;
	 	L =LL;
		R = RR;	
		key = KEY;	
	}
	node()
	{
		ls = rs = this;	
		key = L = R = -1;
	}
}root, Tnull, *null = &Tnull;

void mt(node &now, int LL, int RR)
{
	if (LL == RR)	
	{
		now.key = height[LL];	
		return;
	}
	int mid = (LL + RR) / 2;
	if (now.ls == null)	now.ls = new node(LL, mid, 0, null, null);
	else {
		now.ls -> L = LL;
		now.ls -> R = mid;
	}
	if (now.rs == null)	now.rs = new node(mid + 1, RR, 0, null, null);
	else{
		now.rs -> L = mid + 1;
		now.rs -> R = RR;	
	}
	mt(*now.ls, LL, mid);
	mt(*now.rs, mid + 1, RR);
	int a =now.ls -> key;
	int b = now.rs -> key;
	now.key = min(a, b);	
	return;
}

int find(node &now, int LL, int RR)
{
	if (now.L == LL && now.R == RR)	return	now.key;
	int mid = (now.L + now.R) / 2;
	if (RR <= mid)	return find(*now.ls, LL, RR);
	if (mid < LL)	return find(*now.rs, LL, RR);
	int a = find(*now.ls, LL, mid);
	int b = find(*now.rs, mid + 1, RR);
	return min(a, b);
}

void doit()
{
	root = node(0, totlen, 0, null, null);
	mt(root, 0, totlen);
	int ans = 0;
	for (int i = 2; i <= 2 * textlen; ++ i)
	{
		int a = rank[i];
		int b = rank[totlen - i];
		if (a > b)	swap(a, b);
		int tmp = find(root, a + 1, b) - 1;
		ans = max(ans, tmp );
	}
	cout<<ans<<endl;
}

int main()
{
	int count=0;
	while (1)
	{
		++ count;
		gets(text);	
		if (text[0] == 'E')	break;
		cout<<"Case "<<count<<": ";
		textlen = strlen(text);
		HZSZ(); 
		doit();
	}
	return 0;
}


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