POJ 3581 三段字符串(后缀数组)

Sequence

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 7923		Accepted: 1801
Case Time Limit: 2000MS

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}

Source

POJ Founder Monthly Contest – 2008.04.13, Yao Jinyu

题意：给一个数列，将其分三段，使得每段分别反转组合后的字典序最小。

思路：P381

代码：

//#include"bits/stdc++.h"
#include"cstdio"
#include"map"
#include"set"
#include"cmath"
#include"queue"
#include"vector"
#include"string"
#include"cstring"
#include"ctime"
#include"iostream"
#include"cstdlib"
#include"algorithm"
#define db double
#define ll long long
#define ull unsigned long long
#define vec vector<ll>
#define mt  vector<vec>
#define ci(x) scanf("%d",&x)
#define cd(x) scanf("%lf",&x)
#define cl(x) scanf("%lld",&x)
#define pi(x) printf("%d
",x)
#define pd(x) printf("%f
",x)
#define pl(x) printf("%lld
",x)
//#define rep(i, x, y) for(int i=x;i<=y;i++)
#define rep(i, n) for(int i=0;i<n;i++)
using namespace std;
const int N   = 2e5 + 5;
const int mod = 1e9 + 7;
const int MOD = mod - 1;
const int inf = 0x3f3f3f3f;
const db  PI  = acos(-1.0);
const db  eps = 1e-10;
int sa[N],rev[N];
int r[N];
int tmp[N];
int a[N];
int n,k;
int p1,p2;
bool cmp(int i,int j){
    if(r[i] != r[j]) return r[i]<r[j];
    else
    {
        int ri=i+k<=n?r[i+k]:-1;
        int rj=j+k<=n?r[j+k]:-1;
        return ri<rj;
    }
}
void bulid(int s[N],int l,int *sa)
{

    for(int i=0;i<=l;i++){
        sa[i]=i;
        r[i]=i<l?s[i]:-1;
    }
    for(k=1;k<=l;k*=2){
        sort(sa,sa+l+1,cmp);
        tmp[sa[0]]=0;
        for(int i=1;i<=l;i++){
            tmp[sa[i]]=tmp[sa[i-1]]+(cmp(sa[i-1],sa[i])?1:0);
        }
        for(int i=0;i<=l;i++){
            r[i]=tmp[i];
        }
    }
}
void cal()
{
    memset(rev,0, sizeof(rev));
    memset(r,0, sizeof(r));
    memset(sa,0, sizeof(sa));
    reverse_copy(a,a+n,rev);
    bulid(rev,n,sa);
    for(int i=0;i<=n;i++){
        p1=n-sa[i];
        if(p1>=1&&n-p1>=2){
            break;
        }
    }
    int m=n-p1;
    reverse_copy(a+p1,a+n,rev);
    reverse_copy(a+p1,a+n,rev+m);
    bulid(rev,2*m,sa);
    for(int i=0;i<=2*m;i++){
        p2=p1+m-sa[i];
        if(p2-p1>=1&&n-p2>=1){
            break;
        }
    }
    reverse(a,a+p1);
    reverse(a+p1,a+p2);
    reverse(a+p2,a+n);
    for(int i=0;i<n;i++) printf("%d
",a[i]);
}
int main()
{
    ci(n);
    for(int i=0;i<n;i++) ci(a[i]);
    cal();
    return 0;
}