POJ3581:Sequence(后缀数组)

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

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}

题意：把数组分成3分，每一份都翻转。要求输出字典序最小的结果

思路：參照http://blog.163.com/just_gogo/blog/static/1914390652011823103842787/

第二次求之所以要复制一次我们能够这样理解

首先320505。单纯这个的话，后缀按字典序排，第一是05，第二是0505

可是复制了之后，是第一是0505320505，第二是05320505

可见复制一遍还是有必要的

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;

#define LS 2*i
#define RS 2*i+1
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 2000005
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EXP 1e-8
int wa[N],wb[N],wsf[N],wv[N],sa[N];
int rank[N],height[N],s[N];
//sa:字典序中排第i位的起始位置在str中第sa[i]
//rank:就是str第i个位置的后缀是在字典序排第几
//height：字典序排i和i-1的后缀的最长公共前缀
int cmp(int *r,int a,int b,int k)
{
    return r[a]==r[b]&&r[a+k]==r[b+k];
}
void getsa(int *r,int *sa,int n,int m)//n要包括末尾加入的0
{
    int i,j,p,*x=wa,*y=wb,*t;
    for(i=0; i<m; i++)  wsf[i]=0;
    for(i=0; i<n; i++)  wsf[x[i]=r[i]]++;
    for(i=1; i<m; i++)  wsf[i]+=wsf[i-1];
    for(i=n-1; i>=0; i--)  sa[--wsf[x[i]]]=i;
    p=1;
    j=1;
    for(; 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++)  wsf[i]=0;
        for(i=0; i<n; i++)  wsf[wv[i]]++;
        for(i=1; i<m; i++)  wsf[i]+=wsf[i-1];
        for(i=n-1; i>=0; i--)  sa[--wsf[wv[i]]]=y[i];
        t=x;
        x=y;
        y=t;
        x[sa[0]]=0;
        for(p=1,i=1; i<n; i++)
            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)? p-1:p++;
    }
}

struct node
{
    int id,num;
} a[N];

int cmp1(node a,node b)
{
    if(a.num!=b.num)
        return a.num<b.num;
    return a.id<b.id;
}

int main()
{
    int n,i,j,k,maxn;
    scanf("%d",&n);
    for(i = n-1; i>=0; i--)
    {
        scanf("%d",&a[i].num);
        a[i].id = i;
    }
    sort(a,a+n,cmp1);
    for(i = 0; i<n; i++)
    {
        if(i && a[i].num == a[i-1].num)
        {
            s[a[i].id] = s[a[i-1].id];
            continue;
        }
        s[a[i].id] = i+1;
    }
    s[n] = 0;
    getsa(s,sa,n+1,n+10);
    for(i = 1; i<=n&&sa[i]<=1; i++);

    int tem = sa[i];
    for(i = tem; i<n; i++)
        printf("%d\n",a[s[i]-1].num);
    for(i = 0; i<tem; i++)
        s[i+tem] = s[i];
    tem*=2;
    s[tem] = 0;
    getsa(s,sa,tem+1,n+10);
    int p;
    for(i = 1; i<tem&&(!sa[i]||sa[i]>=tem/2); i++) ;
    p = sa[i];
    for(i = p; i<tem/2; i++)
        printf("%d\n",a[s[i]-1].num);
    for(i = 0; i<p; i++)
        printf("%d\n",a[s[i]-1].num);

    return 0;
}

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