Codeforces 1326D

Description
给一个字符串$s$，从中选出一个前缀与一个后缀（不重叠）前后拼在一起，且拼在一起的串是回文，求最大长度的此类串
Solution
官方的做法是用一个小结论：
设前缀长度为$l$，后缀长度为$r$，
则$forall i in [1,min(l, r)],s[i] = s[len - i + 1]$
设$k = min(l, r)_{max}$, 答案必有一种方案满足$min(l, r) = k$
当$min(l, r) < k$，不妨设$l < r$，则$s[l+1]=s[len - l]=s[len - r + 1]$
我们可以让$r--，l++$，这样仍是长度相同的回文串
于是我们先找出$k$，然后把前$k$个和后$k$个字符删除
再找出最长前缀回文和最长后缀回文比较选择
找出最长前缀回文最好想的是$hash$(用$map$好像会$MLE$)
标程是将原串与反串连在一起，中间加一个非小写字母的字符，然后跑一遍$kmp$

当然还有更弱智的做法：
这题是多组数据，考场我用$map$没清零导致爆炸，就开始质疑上面那个正确的结论$dotsdots$
然后就发现这是一道回文自动机裸题，求出以每个位置为结尾或开头的最长回文串，然后对于$i <= k$，它的贡献就是$2 * i +$以$i+1$开头的最长回文串/以$len-i$结尾的最长回文串，当然一旦选取最长回文串使得前后缀重叠就要跳$fail$链，但这样复杂度无法保证$O(n)$
于是我们不需要每次都求最长回文串，每次只要求到不导致重叠的最长回文串就行

我的两份代码：

#include <bits/stdc++.h>
 
using namespace std;
 
 
int T, len, a, b, nxt[2000010];
 
char s[1000010];
 
int solve(string ss) {
    string t = ss;
    reverse(ss.begin(), ss.end());
    t = ' ' + t + '#' + ss;
    //cerr << t << "-------------------
";
    int j = 0;
    for (int i = 2; i < t.size(); i++) {
        while (j > 0 && t[i] != t[j + 1]) j = nxt[j];
        if (t[i] == t[j + 1]) j++;
        nxt[i] = j;
    }
    return j;
}
 
 
int main() {
    cin >> T;
    while (T--) {
        scanf ("%s", s + 1);
        len = strlen(s + 1);
        a = 1, b = len;
        for (a = 1, b = len; a < b && s[a] == s[b]; a++, b--);
        if (a >= b) {
            for (int i = 1; i <= len; i++) putchar(s[i]);
            putchar('
');
            continue;
        }
        string str = "";
        for (int i = a; i <= b; i++) str += s[i];
        int ans1 = solve(str);
        str = "";
        for (int i = b; i >= a; i--) str += s[i];
        int ans2 = solve(str);
        //cout << a << " " << b << " " << ans1 << " " << ans2 << endl;
        if (ans1 < ans2) {
            for (int i = 1; i < a; i++) putchar(s[i]);
            for (int i = b - ans2 + 1; i <= len; i++) putchar(s[i]);
        }
        else {
            for (int i = 1; i < a + ans1; i++) putchar(s[i]);
            for (int i = b + 1; i <= len; i++) putchar(s[i]);
        }
        putchar('
');
    }
}

#include <bits/stdc++.h>
using namespace std;
 
 
int T, siz, tot, lst, fail[1000010], len[1000010], trans[1000010][26];
 
int ans1, ans2, pos1, pos2;
 
char s[1000010], t[1000010];
 
 
void Insert1(int c, int N) {
    int cur = lst;
    while (t[N - len[cur] - 1] != t[N]) cur = fail[cur];
    int now = trans[cur][c];
    if (!now) {
        now = ++tot;
        memset(trans[now], 0, sizeof(trans[now]));
        int v = fail[cur];
        while (t[N - len[v] - 1] != t[N]) v = fail[v];
        len[now] = len[cur] + 2;
        fail[now] = trans[v][c];
        trans[cur][c] = now;
        //num[now] = num[fail[now]] + 1;
    }
    lst = now;
}
 
void Insert2(int c, int N) {
    int cur = lst;
    while (s[N - len[cur] - 1] != s[N]) cur = fail[cur];
    int now = trans[cur][c];
    if (!now) {
        now = ++tot;
        memset(trans[now], 0, sizeof(trans[now]));
        int v = fail[cur];
        while (s[N - len[v] - 1] != s[N]) v = fail[v];
        len[now] = len[cur] + 2;
        fail[now] = trans[v][c];
        trans[cur][c] = now;
        //num[now] = num[fail[now]] + 1;
    }
    lst = now;
}
 
 
int main() {
    cin >> T;
    while (T--) {
        scanf("%s", s + 1);
        siz = strlen(s + 1);
        int k;
        for (k = 1; k <= siz - k && s[k] == s[siz - k + 1]; k++);
        if (k > siz - k) {
            for (int i = 1; i <= siz; i++) putchar(s[i]);
            putchar('
');
            continue;
        }
        ans1 = ans2 = pos1 = pos2 = 0;
        for (int i = 0; i <= siz; i++) fail[i] = 0;
        fail[1] = 1, fail[0] = 1, len[1] = -1, len[0] = lst = 0, tot = 1;
        memset(trans[0], 0, sizeof(trans[0]));
        memset(trans[1], 0, sizeof(trans[1]));
        for (int i = 1; i <= siz; i++) {
            t[i] = s[siz - i + 1];
            Insert1(t[i] - 'a', i);
            if (i + k > siz) {
                while (lst > 1 && siz - i + len[lst] > i) lst = fail[lst];
                if (ans1 < 2 * (siz - i) + len[lst]) {
                    ans1 = 2 * (siz - i) + len[lst];
                    pos1 = siz - i;
                }
            }
        }
        for (int i = 0; i <= siz; i++) fail[i] = 0;
        fail[1] = 1, fail[0] = 1, len[1] = -1, len[0] = lst = 0, tot = 1;
        memset(trans[0], 0, sizeof(trans[0]));
        memset(trans[1], 0, sizeof(trans[1]));
        for (int i = 1; i <= siz; i++) {
            Insert2(s[i] - 'a', i);
            if (i + k > siz) {
                while (lst > 1 && siz - i + len[lst] > i) lst = fail[lst];
                if (ans2 < 2 * (siz - i) + len[lst]) {
                    ans2 = 2 * (siz - i) + len[lst];
                    pos2 = siz - i;
                }
            }
        }
        if (ans1 < ans2) {
            for (int i = 1; i <= pos2; i++) putchar(s[i]);
            for (int i = siz - (ans2 - pos2) + 1; i <= siz; i++) putchar(s[i]);
        }
        else {
            for (int i = 1; i <= ans1 - pos1; i++) putchar(s[i]);
            for (int i = siz - pos1 + 1; i <= siz; i++) putchar(s[i]);
        }
        putchar('
');
    }
    return 0;
}