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\n",x)
 #define pd(x) printf("%f\n",x)
 #define pl(x) printf("%lld\n",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 + ;
 const int mod = 1e9 + ;
 const int MOD = mod - ;
 const int inf = 0x3f3f3f3f;
 const db  PI  = acos(-1.0);
 const db  eps = 1e-;
 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]:-;
         int rj=j+k<=n?r[j+k]:-;
         return ri<rj;
     }
 }
 void bulid(int s[N],int l,int *sa)
 {

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