題意:求最長回文子串
這種有專門的O(n)套板子算法,但作為練習還是用後綴數組來解吧
只需把相同的另一個串反接(中間用一個足夠小且未出現的字符銜接),然後枚舉回文串的中點,不斷求解該點往前和往後計算的\(LCP\)即可
發現模板有個BUG改好了
有個值得註意的地方是回文長度奇偶枚舉時的端點選擇問題,具體直接看栗子
abcccd
奇數枚舉時應該是abcccd#dcccba
偶數枚舉時應該是abcccd#dcccba
兩個子串枚舉首端與#的距離相等或相差一,列出式子就是當奇數端枚舉\(i\)時,另一子串始端為\(2n+2-i\),偶數端為\(2n+3-i\)

#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;
}

URAL - 1297 後綴數組的做法 LCP應用