POJ 3581-Sequence(后缀数组)

Sequence

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 6485		Accepted: 1429
Case Time Limit: 2000MS

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}
Source

POJ Founder Monthly Contest – 2008.04.13, Yao Jinyu

题目意思：

有一个数组，分割成三份，每份翻转后，得到的最小的字典序?

解题思路：

后缀数组。先确定第一段，因为A1是最大的，所以求反转之后字符串中字典序最小的后缀。

再将剩余部分分成两段，计算两个原序列并将得到的新序列反转后的后缀数组，从中选取字典序最小的合适的后缀。

cin超时，输入输出外挂WA，scanf才AC…坎坷…

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
typedef long long ll;
#define maxn 200100

int n,k,a[maxn];
int rev[maxn],sa[maxn];
int rank[maxn*2],tmp[maxn*2];

bool compare_sa(int i,int j)//倍增法，比较rank
{
if(rank[i]!=rank[j]) return rank[i]<rank[j];
else
{
int ri=i+k<=n?rank[i+k] :-1;
int rj=j+k<=n?rank[j+k] :-1;
return ri<rj;
}
}

void construct_sa(int s[],int N,int sa[])//计算s的后缀数组
{
for(int i=0; i<=N; ++i)//初始长度为1，rank为字符编码
{
sa[i]=i;
rank[i]=i<N?s[i] :-1;
}
for(k=1; k<=N; k*=2)//倍增法求后缀数组
{
sort(sa,sa+N+1,compare_sa);
tmp[sa[0]]=0;
for(int i=1; i<=N; ++i)
tmp[sa[i]]=tmp[sa[i-1]]+(compare_sa(sa[i-1],sa[i])?1:0);
for(int i=0; i<=N; ++i)
rank[i]=tmp[i];
}
}

int main()
{
ios::sync_with_stdio(false);
cin>>n;
for(int i=0; i<n; ++i)
cin>>a[i];
//solve();
reverse_copy(a,a+n,rev);//a不变，反转a之后存入rev数组
construct_sa(rev,n,sa);//求后缀数组
int p1;
for(int i=0; i<n; ++i)//确定第一段的分割位置
{
p1=n-sa[i];
if((p1>=1)&&n-p1>=2) break;
}
//cout<<"p1="<<p1<<endl;
int m=n-p1;
reverse_copy(a+p1,a+n,rev);
reverse_copy(a+p1,a+n,rev+m);
construct_sa(rev,m*2,sa);
int p2;
for(int i=0; i<=2*m; ++i)//确定第二段的分割位置
{
p2=p1+m-sa[i];
if((p2-p1>=1)&&n-p2>=1) break;
}
//cout<<"p2="<<p2<<endl;
reverse(a,a+p1);
reverse(a+p1,a+p2);
reverse(a+p2,a+n);
for(int i=0; i<n; ++i)
cout<<a[i]<<endl;
return 0;
}

/**
5
10
1
2
3
4**/