Suffix array tyvj P1860 suffix array

/*p1860 suffix Array
Time: 1000ms/Space: 131072kib/java class Name: Main
Describe

We define a string with a suffix of suffix (i) representing the substring from s[i] to S[length (s)].
The suffix array (Suffix array) Sa[i] holds an arrangement that satisfies Suffix (Sa[i]) <suffix (sa[i+1]) compared in dictionary order
Defines Height[i] represents the longest common prefix length between suffix (sa[i]) and suffix (sa[i-1]), where height[1]=0
Your task is to find the two arrays of SA and height. String length <=200000
Input format

A line, which is a string in the description (only lowercase letters will appear)
Output format

A total of two lines, n number per row, first behavior Sa[i], second behavior height[i], where each line is separated by a space
Test Sample 1

Input

Aabaaaab
Output

4 5 6 1 7 2 8 3
0 3 2 3 1 2 0 1*/
The topic algorithm is the title name, but also is the most simple suffix array application.
About suffix arrays I don't want to say more, recommended algorithm collection of "suffix array-a powerful tool for handling strings"
Sometimes I really want someone to tell me that self-study is very, very painful.
But 2 points timed out, but no matter what.
#include <cstdio>
#include <iostream>
#include <cstring>
#define N 200008
using namespace Std;
int n,sa[2][n],rk[2][n],a[n],v[n],k,h[n];
Char Ch[n];
void Geng (int sa[n],int rk[n],int sa[n],int rk[n])
{
for (int i=1;i<=n;i++)
V[rk[sa[i]]]=i;
for (int i=n;i;i--)
if (sa[i]>k)
Sa[v[rk[sa[i]-k]]--]=sa[i]-k;
for (int i=n-k+1;i<=n;i++)
Sa[v[rk[i]]--]=i;
for (int i=1;i<=n;i++)
rk[sa[i]]=rk[sa[i-1]]+ (rk[sa[i]]!=rk[sa[i-1]]| | Rk[sa[i]+k]!=rk[sa[i-1]+k]);
Return
}
int main ()
{
Gets (ch+1);
N=strlen (ch+1);
for (int i=1;i<=n;i++)
a[i]=ch[i]-' a ' +1;
int p=0,q=1;
for (int i=1;i<=n;i++)
v[a[i]]++;
for (int i=2;i<27;i++)
V[I]+=V[I-1];
for (int i=1;i<=n;i++)
Sa[p][v[a[i]]--]=i;
for (int i=1;i<=n;i++)
rk[p][sa[p][i]]=rk[p][sa[p][i-1]]+ (A[sa[p][i]]!=a[sa[p][i-1]);
for (K=1;k<n;k<<=1)
{
Geng (Sa[p],rk[p],sa[q],rk[q]);
Swap (P,Q);
}
for (int i=1;i<=n;i++)
printf ("%d", sa[p][i]);
printf ("\ n");
k=0;
for (int i=1;i<=n;i++)
if (rk[p][i]==1)
h[1]=0;
Else
{
int j=sa[p][rk[p][i]-1];
for (; a[i+k]==a[j+k];k++);
H[rk[p][i]]=k;
if (k>0)
k--;
}
for (int i=1;i<=n;i++)
printf ("%d", h[i]);
return 0;
}

Suffix array tyvj P1860 suffix array