UVA 1297 最长回文串【裸】

题目大意：给一行字母，只有字母，最多1000个字母。 求最长回文穿，有多个情况下，输出最前面的那个。

这次终于让我后缀数组不TLE了…… 哇哇哇哇哇！  我就是喜欢用线段树写RMQ……慢就慢，我不管了

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


const int max_strlen = 4000  + 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;
	for (int i = 0; i != textlen; ++ i)	R[totlen ++] = text[i];
	R[totlen ++] = 1; //串分割符号
	for (int i = textlen - 1; i >=0 ; -- i)	R[totlen ++] = text[i];
	R[totlen] = 0; //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;

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

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 = 1, flag = 1 ,wz = 0;
	for (int i = 1; i != textlen; ++ i)
	{
		int a = rank[i];
		int b = rank[totlen - i - 1];
		if (a > b)	swap(a, b);
		int tmp = (find(root, a + 1, b) * 2) - 1;
		if (tmp> ans)	
		{
			ans = tmp;;
			flag = 1;	
			wz = i;
		}
		a = rank[i];
		b = rank[totlen - i];
		if (a > b)	swap(a, b);
		tmp = find(root, a + 1, b) * 2;
		if (tmp > ans)
		{
			ans = tmp;
			flag = 0;	
			wz = i;
		}
	}
	if (flag == 1) //奇数的情况
	{
		int tmp = ans / 2;
		for (int i = wz - tmp; i < wz + tmp; ++ i)	cout<<text[i];
		cout<<text[wz + tmp];
	}else{
		int tmp = ans / 2;
		for (int i = wz - tmp; i < wz + tmp - 1; ++ i)	cout<<text[i];
		cout<<text[wz + tmp - 1];	
	}
}

int main()
{
	gets(text);	
	textlen = strlen(text);
	HZSZ(); 
	doit();
	return 0;
}