POJ

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

5
10
1
2
3
4

Sample Output

1
10
2
4
3

题意:
分成三个非空段,分别翻转,求字典序最小的方案.
思路:
将其翻转,找到字典序最小的后缀,要求起点在[0,n-2) 内,即可找到第一段.
第一段去除之后,对原串翻转,并复制,找到字典序最小的,即可找到第二段和第三段.
假设原串是 ab
翻转复制后:b'a'b'a'
a'b'就是最终的结果,所以这样做是正确的.
当然要注意后缀起点位置.

#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>

#define fuck(x) cerr<<#x<<" = "<<x<<endl;
#define debug(a, x) cerr<<#a<<"["<<x<<"] = "<<a[x]<<endl;
#define ls (t<<1)
#define rs ((t<<1)|1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 200086;
const int maxm = 100086;
const int inf = 0x3f3f3f3f;
const ll Inf = 999999999999999999;
const int mod = 1000000007;
const double eps = 1e-6;
const double pi = acos(-1);

int s[maxn];
int len, Rank[maxn], sa[maxn], tlen, tmp[maxn];

bool compare_sa(int i, int j) {
    if (Rank[i] != Rank[j]) { return Rank[i] < Rank[j]; }
    //如果以i开始,长度为k的字符串的长度,已经超出了字符串尾,那么就赋值为-1
    //这是因为,在前面所有数据相同的情况下,字符串短的字典序小.
    int ri = i + tlen <= len ? Rank[i + tlen] : -inf;
    int rj = j + tlen <= len ? Rank[j + tlen] : -inf;
    return ri < rj;
}

void construct_sa() {
    //初始的RANK为字符的ASCII码
    for (int i = 0; i <= len; i++) {
        sa[i] = i;
        Rank[i] = i < len ? s[i] : -inf;
    }
    for (tlen = 1; tlen <= len; tlen *= 2) {
        sort(sa, sa + len + 1, compare_sa);
        tmp[sa[0]] = 0;
        //全新版本的RANK,tmp用来计算新的rank
        //将字典序最小的后缀rank计为0
        //sa之中表示的后缀都是有序的,所以将下一个后缀与前一个后缀比较,如果大于前一个后缀,rank就比前一个加一.
        //否则就和前一个相等.
        for (int i = 1; i <= len; i++) {
            tmp[sa[i]] = tmp[sa[i - 1]] + (compare_sa(sa[i - 1], sa[i]) ? 1 : 0);
        }
        for (int i = 0; i <= len; i++) {
            Rank[i] = tmp[i];

        }
    }
}

int height[maxn];

void construct_lcp() {
//    for(int i=0;i<=n;i++){Rank[sa[i]]=i;}
    int h = 0;
    height[0] = 0;
    for (int i = 0; i < len; i++) {//i为后缀数组起始位置
        int j = sa[Rank[i] - 1];//获取当前后缀的前一个后缀(排序后)
        if (h > 0)h--;
        for (; j + h < len && i + h < len; h++) {
            if (s[j + h] != s[i + h])break;
        }
        height[Rank[i]] = h;
    }
}

int st[maxn][20];

void rmq_init() {
    for (int i = 1; i <= len; i++) {
        st[i][0] = height[i];
    }
    int l = 2;
    for (int i = 1; l <= len; i++) {
        for (int j = 1; j + l / 2 <= len; j++) {
            st[j][i] = min(st[j][i - 1], st[j + l / 2][i - 1]);
        }
        l <<= 1;
    }
}

int ask_min(int i, int j) {
    int k = int(log(j - i + 1.0) / log(2.0));
    return min(st[i][k], st[j - (1 << k) + 1][k]);
}

int lcp(int a, int b)//此处参数是,原字符串下标
{
    a = Rank[a], b = Rank[b];
    if (a > b)
        swap(a, b);
    return ask_min(a + 1, b);
}

int solve(int l, int r) {
    int ans = 0;
    for (int i = 1; i <= len; i++) {
//        fuck(sa[i])
        if (sa[i] >= l && sa[i] <= r) {
            ans = sa[i];
            break;
        }
    }
    return ans;
}

int num[maxn];

int main() {
//    ios::sync_with_stdio(false);
//    freopen("in.txt", "r", stdin);
    int n;
    scanf("%d", &n);
    len = n;
    for (int i = 0; i < n; i++) {
        scanf("%d", &s[i]);
        num[i] = s[i];
    }
    reverse(s, s + n);
    construct_sa();
    int fi = solve(2, n - 1);
    fi = len - fi - 1;
    reverse(num, num + fi + 1);

    len = 0;
    for (int i = fi + 1; i < n; i++) {
        s[len++] = num[i];
    }
    reverse(s, s + len);
    for(int i=0;i<len;i++){
        s[i+len]=s[i];
    }len*=2;
    construct_sa();

    int se = solve(1, len /2-1);

    se = n - se - 1;
    reverse(num + fi + 1, num + se + 1);
    reverse(num + se + 1, num + n);
    for (int i = 0; i < n; i++) {
        printf("%d
", num[i]);
    }
    return 0;
}

Sequence

 POJ - 3581