POJ3581:Sequence(后缀数组)
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}
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：参照 http://blog.163.com/just_gogo/blog/static/1914390652011823103842787/
第二次求之所以要复制一次我们可以这样理解
首先320505，单纯这个的话，后缀按字典序排，第一是05，第二是0505
但是复制了之后，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要包含末尾添加的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;
}