hdu3068最长回文 后缀数组TLE版本23333

感觉学习了后缀数组整个人都好了，于是看到这题想都没想就开始写，然而事实并不如意，TTTTTTTTQAQQQQQQ，赶快吧O复杂度换成T复杂度，真是玄学4亿多= =

放下个TLE的版本，还有就是被之前学的版本坑了，在原串和反串之间加入了特殊符号之后，height数组的最大值就已经是答案了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#define LL long long
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define down(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
#define N 220005
#define M 30
char s
;
int sa
,height
,ra
;
int n,ans;

int C
;
void getsa(int m)
{
fo(i,0,n)ra[i]=0,sa[i]=0,height[i]=0;
fo(i,0,m)C[i]=0;
fo(i,1,n)ra[i]=s[i]-96,C[ra[i]]++;
fo(i,1,m)C[i]+=C[i-1];
// cout<<n<<' '<<ra
<<' '<<C[ra
]<<' '<<sa[C[ra
]]<<endl;
down(i,n,1)sa[C[ra[i]]--]=i;
for(int k=1;k<=n;k<<=1)
{
int p=0;
fo(i,n-k+1,n)height[++p]=i;
fo(i,1,n)if(sa[i]>k)height[++p]=sa[i]-k;
fo(i,0,m)C[i]=0;
fo(i,1,n)C[ra[height[i]]]++;
fo(i,1,m)C[i]+=C[i-1];
down(i,n,1)sa[C[ra[height[i]]]--]=height[i];

swap(ra,height);p=0;ra[sa[1]]=++p;
fo(i,2,n)
if(height[sa[i]]==height[sa[i-1]]&&height[sa[i]+k]==height[sa[i-1]+k])
ra[sa[i]]=p;else ra[sa[i]]=++p;//cout<<"Run:"<<k<<endl;
if(n==p)break;
m=p;
}
// fo(i,1,n)fo(j,sa[i],n)cout<<s[j];cout<<endl;
}

void getheight()
{
int i=0,j=0,k=0;
for(i=1;i<=n;height[ra[i++]]=k,ans=max(ans,k))
for(k?k--:0,j=sa[ra[i]-1];s[i+k]==s[j+k];k++);
// fo(i,1,n)cout<<height[i]<<' ';cout<<endl;
}

int main()
{
while(scanf("%s",s+1)!=EOF)
{
n=strlen(s+1);int len=n;s[n+1]='z'+1;ans=0;
for(int i=1,j=2*n+1;i<j;i++,j--)s[j]=s[i];n<<=1;n++;
getsa(27);
getheight();
// fo(i,2,n)ans=max(height[i],ans);
printf("%d\n",ans);
}
return 0;
}