题意:求最长回文子串
这种有专门的O(n)套板子算法,但作为练习还是用后缀数组来解吧
只需把相同的另一个串反接(中间用一个足够小且未出现的字符衔接),然后枚举回文串的中点,不断求解该点往前和往后计算的\(LCP\)即可
发现模板有个BUG改好了
有个值得注意的地方是回文长度奇偶枚举时的端点选择问题,具体直接看栗子
abcccd
奇数枚举时应该是abcccd#dcccba
偶数枚举时应该是abcccd#dcccba
两个子串枚举首端与#的距离相等或相差一,列出式子就是当奇数端枚举\(i\)时,另一子串始端为\(2n+2-i\),偶数端为\(2n+3-i\)
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define iin(a) scanf("%d",&a)
#define lin(a) scanf("%lld",&a)
#define din(a) scanf("%lf",&a)
#define s0(a) scanf("%s",a)
#define s1(a) scanf("%s",a+1)
#define print(a) printf("%lld",(ll)a)
#define enter putchar(‘\n‘)
#define blank putchar(‘ ‘)
#define println(a) printf("%lld\n",(ll)a)
#define IOS ios::sync_with_stdio(0)
using namespace std;
const int maxn = 1e5+11;
const int oo = 0x3f3f3f3f;
const double eps = 1e-7;
typedef long long ll;
ll read(){
ll x=0,f=1;register char ch=getchar();
while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
return x*f;
}
char str[maxn];int n;
struct SA{
int Rank[maxn],sa[maxn],tsa[maxn],A[maxn],B[maxn];
int cntA[maxn],cntB[maxn];
int height[maxn],best[maxn][30],n;
void get(int nn){
n=nn;
rep(i,0,666) cntA[i]=0;
rep(i,1,n) cntA[str[i]]++;
rep(i,1,666) cntA[i]+=cntA[i-1];
rrep(i,n,1) sa[cntA[str[i]]--]=i;
Rank[sa[1]]=1;
rep(i,2,n){
if(str[sa[i]]==str[sa[i-1]]){
Rank[sa[i]]=Rank[sa[i-1]];
}else{
Rank[sa[i]]=1+Rank[sa[i-1]];
}
}
for(int l=1;Rank[sa[n]]<n;l<<=1){
rep(i,1,n) cntA[i]=cntB[i]=0;
rep(i,1,n) cntA[A[i]=Rank[i]]++;
rep(i,1,n) cntB[B[i]=(i+l<=n?Rank[i+l]:0)]++;
rep(i,1,n) cntA[i]+=cntA[i-1],cntB[i]+=cntB[i-1];
rrep(i,n,1) tsa[cntB[B[i]]--]=i;
rrep(i,n,1) sa[cntA[A[tsa[i]]]--]=tsa[i];
Rank[sa[1]]=1;
rep(i,2,n){
bool flag=A[sa[i]]==A[sa[i-1]]&&B[sa[i]]==B[sa[i-1]];
flag=!flag;
Rank[sa[i]]=Rank[sa[i-1]]+flag;
}
}
}
void ht(){
int j=0;
rep(i,1,n){
if(j) j--;
while(str[i+j]==str[sa[Rank[i]-1]+j]) j++;
height[Rank[i]]=j;
}
}
void rmq(){
rep(i,1,n) best[i][0]=height[i];
for(int i=1;(1<<i)<=n;i++){
for(int j=1;j+(1<<i)-1<=n;j++){
best[j][i]=min(best[j][i-1],best[j+(1<<(i-1))][i-1]);
}
}
}
int query(int l,int r){
if(l==r)return -oo;
if(l>r)swap(l,r);
l++;
int k=log2(r-l+1);
return min(best[l][k],best[r-(1<<k)+1][k]);
}
}sa;
char sstr[maxn];
int main(){
while(~s1(str)){
n=strlen(str+1);
rep(i,1,n+1) sstr[i]=str[i];
reverse(sstr+1,sstr+1+n);
str[n+1]=1;str[n+2]=0;
strcat(str+1,sstr+1);
str[2*n+2]=0;
int nn=2*n+1;
sa.get(nn);
sa.ht();
sa.rmq();
int ans=1,pos=1;
rep(i,1,n){
if(1){
int odd=sa.query(sa.Rank[i],sa.Rank[2*n+2-i]);
if(ans<2*odd-1)ans=2*odd-1,pos=i-odd+1;
}
int even=sa.query(sa.Rank[i],sa.Rank[2*n+3-i]);
if(ans<2*even)ans=2*even,pos=i-even;
}
// cout<<ans<<endl;
str[ans+pos]=0;printf("%s\n",str+pos);
memset(str,0,sizeof str);
memset(sstr,0,sizeof sstr);
}
return 0;
}