POJ 3581 Sequence

Description

Given a sequence, {A1, A2, ..., An} which is guaranteed A1 > A2, ..., An,  you are to cut it into three sub-sequences and reverse them
separately to form a new one which is the smallest possible sequence in alphabet order.

The alphabet order is defined as follows: for two sequence {A1, A2, ..., An} and {B1, B2, ..., Bn}, we say {A1, A2,
..., An} is smaller than {B1, B2, ..., Bn} if and only if there exists such i ( 1 ≤ i ≤ n) so that we have Ai < Bi and Aj = Bj for
each j < i.

Input

The first line contains n. (n ≤ 200000)

The following n lines contain the sequence.

Output

output n lines which is the smallest possible sequence obtained.

Sample Input

5
10
1
2
3
4

Sample Output

1
10
2
4
3

Hint

{10, 1, 2, 3, 4} -> {10, 1 | 2 | 3, 4} -> {1, 10, 2, 4, 3}

把序列分成非空的三段，每段反转再合并，问字典序最小是多少。
因为A[1]是最大的，所以第一段很简单，求序列反转以后最小的后缀就行了，注意，因为非空，这个后缀不能是第一个或第二个开头的。
然后就是第二段和第三段，在反转的序列中，第二段和第三段就相当于把序列一分为二，位置反过来合并，
这个就是序列的最小表示，可以用最小表示法搞定也可以把序列翻倍再用后缀数组，同样要注意非空的问题。
另外，此题多组数据似乎有问题？
#include<set>
#include<map>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define per(i,j,k) for (int i = j; i >= k; i--)
#define loop(i,j,k) for (int i = j;i != -1; i = k[i])
#define lson x << 1, l, mid
#define rson x << 1 | 1, mid + 1, r
#define fi first
#define se second
#define mp(i,j) make_pair(i,j)
#define pii pair<string,string>
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const double eps = 1e-8;
const int INF = 0x7FFFFFFF;
const int mod = 1e8;
const int N = 5e5 + 10;
const int read()
{
char ch = getchar();
while (ch<'0' || ch>'9') ch = getchar();
int x = ch - '0';
while ((ch = getchar()) >= '0'&&ch <= '9') x = x * 10 + ch - '0';
return x;
}

struct Sa
{
int s
, a
;
int rk[2]
, sa
, h
, w
, now, n, m;
int rmq
[20], lg
;

bool GetS()
{
n = read();
rep(i, 1, n)
{
scanf("%d", &s[n - i + 1]);
a[i] = s[n - i + 1];
}
sort(a + 1, a + n + 1);
m = unique(a + 1, a + n + 1) - a;
rep(i, 1, n) s[i] = lower_bound(a + 1, a + m, s[i]) - a;
return true;
}

void getsa(int z, int &m)
{
int x = now, y = now ^= 1;
rep(i, 1, z) rk[y][i] = n - i + 1;
for (int i = 1, j = z; i <= n; i++)
if (sa[i] > z) rk[y][++j] = sa[i] - z;

rep(i, 1, m) w[i] = 0;
rep(i, 1, n) w[rk[x][rk[y][i]]]++;
rep(i, 1, m) w[i] += w[i - 1];
per(i, n, 1) sa[w[rk[x][rk[y][i]]]--] = rk[y][i];
for (int i = m = 1; i <= n; i++)
{
int *a = rk[x] + sa[i], *b = rk[x] + sa[i - 1];
rk[y][sa[i]] = *a == *b&&*(a + z) == *(b + z) ? m - 1 : m++;
}
}

void getsa(int m)
{
//n = strlen(s + 1);
rk[1][0] = now = sa[0] = s[0] = 0;
rep(i, 1, m) w[i] = 0;
rep(i, 1, n) w[s[i]]++;
rep(i, 1, m) rk[1][i] = rk[1][i - 1] + (bool)w[i];
rep(i, 1, m) w[i] += w[i - 1];
rep(i, 1, n) rk[0][i] = rk[1][s[i]];
rep(i, 1, n) sa[w[s[i]]--] = i;

rk[1][n + 1] = rk[0][n + 1] = 0; //多组的时候容易出bug
for (int x = 1, y = rk[1][m]; x <= n && y <= n; x <<= 1) getsa(x, y);
for (int i = 1, j = 0; i <= n; h[rk[now][i++]] = j ? j-- : j)
{
if (rk[now][i] == 1) continue;
int k = n - max(sa[rk[now][i] - 1], i);
while (j <= k && s[sa[rk[now][i] - 1] + j] == s[i + j]) ++j;
}
}

void getrmq()
{
h[n + 1] = h[1] = lg[1] = 0;
rep(i, 2, n) rmq[i][0] = h[i], lg[i] = lg[i >> 1] + 1;
for (int i = 1; (1 << i) <= n; i++)
{
rep(j, 2, n)
{
if (j + (1 << i) > n + 1) break;
rmq[j][i] = min(rmq[j][i - 1], rmq[j + (1 << i - 1)][i - 1]);
}
}
}

int lcp(int x, int y)
{
int l = min(rk[now][x], rk[now][y]) + 1, r = max(rk[now][x], rk[now][y]);
return min(rmq[l][lg[r - l + 1]], rmq[r - (1 << lg[r - l + 1]) + 1][lg[r - l + 1]]);
}

void work()
{
GetS(); getsa(m);
rep(i, 1, n) if (sa[i] > 2)
{
rep(j, sa[i], n) printf("%d\n", a[s[j]]);
n = sa[i] - 1;
rep(j, 1, n) s[j + n] = s[j];
n <<= 1; getsa(m);
rep(j, 1, n) if (sa[j] > 1 && sa[j] * 2 <= n)
{
rep(k, sa[j], n / 2) printf("%d\n", a[s[k]]);
rep(k, 1, sa[j] - 1) printf("%d\n", a[s[k]]);
break;
}
break;
}
}
}sa;

int main()
{
sa.work();
return 0;
}