題意

給定一個字符串 \(S\)
求所有的 \(S[i,n-i+1]\) 的 \(border\) 長度(最長的前綴等於後綴)，要求長度是奇數
\(n\le 10^6\)

Sol

首先發現每次求的串都是原串去掉前後 \(i-1\) 位得到的串
一個套路，把串翻折，又因為 \(border\) 長度可能大於一半，所以我們把串倍長後翻折
也就是翻轉過來隔空插入在一起
例如：
串 \(bcabcabcabcabca\)
翻轉後 \(acbacbacbacbacb\)
隔一個插入在一起 \(baccabbaccabbaccabbaccabbaccab\)
那麽也就是求這個串的以某個位置的開始的最長回文串

註意到每次都要跳 \(fail\) 鏈跳到滿足要求的位置，而每次都跳很耗時
如果之後跳到之前跳到過的點，就可以直接跳到之前跳到的對答案有貢獻的點上
再繼續跳
並查集維護一下就好了

# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
 

using namespace std;
typedef long long ll;

IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3
 
) + (c ^ 48);
    return x * z;
}

const int maxn(2e6 + 5);

int n, last, tot, anc[maxn], id[maxn];
int len[maxn], first[maxn], nxt[maxn], type[maxn], fa[maxn];
char s[maxn], str[maxn];

IL int Son(RG int u, RG int c){
    for(RG int v = first[u]; v; v = nxt[v])
        if(type[v] == c) return v;
    return 0;
}

IL void Link(RG int u, RG int v, RG int c){
    nxt[v] = first[u], first[u] = v, type[v] = c;
}

IL void Extend(RG int pos, RG int c){
    RG int p = last;
    while(s[pos - len[p] - 1] != s[pos]) p = fa[p];
    if(!Son(p, c)){
        RG int np = ++tot, q = fa[p];
        while(s[pos - len[q] - 1] != s[pos]) q = fa[q];
        len[np] = len[p] + 2, fa[np] = Son(q, c);
        Link(p, np, c);
    }
    last = Son(p, c);
}

IL int Find(RG int x){
    return anc[x] == x ? x : anc[x] = Find(anc[x]);
}

int main(){
    Fill(type, -1), n = Input(), scanf(" %s", str + 1);
    for(RG int t = 0, i = 1, j = n; j; ++i, --j)
        s[++t] = str[i], s[++t] = str[j];
    tot = 1, fa[0] = fa[1] = 1, len[1] = -1, n <<= 1, anc[0] = 1;
    for(RG int i = 1; i <= n; ++i) Extend(i, s[i] - 'a'), id[i] = last;
    for(RG int i = 0; i <= tot; ++i) anc[i] = i;
    for(RG int i = 1, m = n >> 1, t = (m + 1) >> 1; i <= t; ++i){
        RG int x = Find(id[n - ((i - 1) << 1)]);
        while(x != 1 && ((len[x] >> 1) >= (m - ((i - 1) << 1)) || len[x] % 4 != 2)) x = anc[x] = Find(fa[anc[x]]);
        printf("%d ", (len[x] % 4 == 2) ? (len[x] >> 1) : -1);
    }
    return 0;
}

CF961F k-substrings