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;

}