USACO以前的1.3.3 [calfflac] 后缀树组方法

后缀数组构造出的height数组，利用RMQ可以解决回文串问题。

后缀数组的构造需要在后面加一个ASCII码很小的东西，回文串中间也加一个特殊字符比如#

aabbaa 构造后就是 aabbaa#aabbaa*

abcd构造后就是 abcd#dcba*

大概就这些笔记了……

然后就是利用height数组的性质来用平衡树解决RMQ。 用为我用倍增法，所以也就用线段树来解决。整体时间复杂度nlogn，不是优秀的算法，但是对于这题而言还可以的。

/*
TASK:calfflac
LANG:C++
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

const int max_n = 40000 + 5;
int wa[max_n], wb[max_n], wv[max_n], tub[max_n];
int R[max_n], sa[max_n], height[max_n], rank[max_n];

string origin_string, son_origin_string;
int son_reflect_origin[max_n];

inline char uppercase_to_lowercase(char &ch)
{
	if ('A' <= ch && ch <= 'Z')	ch -= 'A' - 'a';
	return ch;
}
#define utl(ch)	uppercase_to_lowercase(ch)
inline bool if_letter(char ch)
{
	if ('a' <= ch && ch <= 'z')	return true;
	return false;
}

int 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 = 200)
{
	int *x = wa, *y = wb, *t;//类似于先桶排，然后利用桶离散化，然后继续桶排，然后再离散化……
	int i, j, p, k = 0;
	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; j *= 2, m = p)
	{
		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 - 1], sa[i], j) ? p - 1 : p ++;
	}
	for (i = 0; i != n; ++ i)	rank[sa[i]] = i; 
	for (i = 0; i != n; height[rank[i++]] = k)
		for (k?k--:0,j=sa[rank[i]-1]; r[i+k] == r[j + k]; k ++);
}

struct node
{
	int L,R;
	int max_num, min_num;
	node *left_son, *right_son;
	node(int L, int R)
	{
		this -> L = L;
		this -> R = R;
	}
	node()
	{
		left_son = NULL;
		right_son = NULL;
		max_num = 0x7fffffff;
		min_num = - 0x7fffffff;
	}
}root;

void build_tree(node &root)
{
	if (root.L == root.R)
	{
		root.max_num = root.min_num = height[root.L];	
		return ;
	}
	int mid = (root.L + root.R) / 2;
	root.left_son = new node(root.L, mid);
	root.right_son = new node(mid + 1, root.R);	
	build_tree(*root.left_son);
	build_tree(*root.right_son);
	root.max_num = max(root.left_son -> max_num, root.right_son -> max_num);
	root.min_num = min(root.left_son -> min_num, root.right_son -> min_num);
	return;
}

int find_tree(node &root, int L, int R)
{
	if (root.L == L && root.R == R)	return root.min_num;
	int mid = (root.L + root.R) / 2;
	if (R <= mid)	return find_tree(*root.left_son, L, R);
	if (mid < L)	return find_tree(*root.right_son, L, R);
	int A = find_tree(*root.left_son, L, mid);
	int B = find_tree(*root.right_son, mid + 1, R);
	return min(A, B);
}

int main()
{
	freopen("calfflac.in", "r", stdin);
	freopen("calfflac.out", "w", stdout);
	char ch;
	while ((ch = getchar()) != EOF)	
	{
		origin_string += ch;
		utl(ch);
		if (if_letter(ch))
		{
			son_origin_string += ch;	
			son_reflect_origin[son_origin_string.length() - 1] = origin_string.length() - 1;
		}	
	}
	int T = son_origin_string.length();
	son_origin_string += '#';
	for (int i = son_origin_string.length() - 2; i >= 0; -- i)	son_origin_string += son_origin_string[i];
	for (int i = 0; i != son_origin_string.length(); ++ i)	R[i] = (int)son_origin_string[i];
	son_origin_string += (char)(5);
	da(R, sa, son_origin_string.length());
	root = node(0, son_origin_string.length());
	build_tree(root);
	int ans = -1;
	int st, ed;
	for (int i = 0; i != T; ++ i)	
	{
		int a = rank[i], b = rank[son_origin_string.length() - i - 2];
		if (a > b)	swap(a, b);
		++ a;
		int tmp = find_tree(root, a, b); 	
		int sb = tmp;
		tmp = tmp * 2 - 1;
		if (tmp > ans)	
		{
			ans = tmp;
			st = i - sb + 1;
			ed = i + sb - 1;	
		}
		
		a = rank[i];
		b = rank[son_origin_string.length() - (i - 1) - 2];
		if (a > b)	swap(a, b);
		++a;
		tmp = find_tree(root, a, b);
		sb = tmp;
		tmp *= 2;
		if (tmp > ans)
		{
			ans = tmp;
			st = i - sb;
			ed = i + sb - 1;	
		}
	}
	cout<<ans<<endl;
	for (int i = son_reflect_origin[st]; i <= son_reflect_origin[ed]; ++ i)	putchar(origin_string[i]);
	cout<<endl;
	return 0;

}