POJ 3581Sequence

Description

Given a sequence, {A₁, A₂, ..., A_n} which is guaranteed A₁ > A₂, ..., A_n,  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, ..., A_n} and {B₁, B₂, ..., B_n}, we say {A₁, A₂, ..., A_n} is smaller than {B₁, B₂, ..., B_n} if and only if there exists such i ( 1 ≤ i ≤ n) so that we have A_i < B_{i and} A_j = B_j 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

5101234

Sample Output

110243

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[N], a[N];int rk[2][N], sa[N], h[N], w[N], now, n, m;int rmq[N][20], lg[N];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bugfor (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;}