1326D Prefix-Suffix Palindrome (Hard version) （manacher）

拉两次车，第一次假设最长回文串就在两边

第二次，先使得左右两边判回文，在将剩下的串 在manachar一遍，

比较两种情况取最大值，然后就是输出问题见代码

#include <bits/stdc++.h>
using namespace std;
 
const int maxn = 1e6 + 5;
 
int pArr[maxn << 1], t;
char s[maxn], chaArr[maxn << 1];
 
int manachar(int l, int r, int& flag)
{
    int len = 0;
    chaArr[len++] = '$', chaArr[len++] = '#';
    for (int i = l; i <= r; ++i) chaArr[len++] = s[i], chaArr[len++] = '#';
    chaArr[len] = '\0';
    int R = -1, C = -1, ans = 0;
    for (int i = 0; i < len; ++i)
    {
        pArr[i] = R > i ? min(R - i, pArr[(C << 1) - i]) : 1;
        while (chaArr[i + pArr[i]] == chaArr[i - pArr[i]]) ++pArr[i];
        if (pArr[i] + i > R) R = pArr[i] + i, C = i;
        if (pArr[i] + i > len - 2 && ans < pArr[i]) ans = pArr[i], flag = 0;
        if (i - pArr[i] < 2 && ans < pArr[i]) ans = pArr[i], flag = 1;
    }
    return ans - 1;
}
 
int main()
{
    cin >> t;
    for (int i = 1; i <= t; ++i)
    {
        cin >> s + 1;
        int flag, len = strlen(s + 1);
        int ans = manachar(1, len, flag);
        int a = 1, b = len, flag2;
        while (a <= b && s[a] == s[b]) ++a, --b;
        if (a > b) { puts(s + 1); continue; }
        int res = manachar(a, b, flag2);
        if (ans >= (a - 1) * 2 + res)
        {
            if (flag) s[ans + 1] = '\0', cout << s + 1 << '\n';
            else cout << s + (len + 1 - ans) << '\n';
            continue;
        }
        if (flag2) s[a + res] = '\0', cout << s + 1 << s + (b + 1) << '\n';
        else s[a] = '\0', cout << s + 1 << s + (b + 1 - res) << '\n';
    }
    return 0;
}