正题

luogu 4555

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题目大意

给出一个字符串，让你求两个相邻的回文串的最大长度和

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解题思路

先用Manacher求出最长回文串

然后暴力配对即可

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代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 200021
using namespace std;
int n, ans, l[N], r[N], v[N], s[N];
string str;
void Manacher()
{int mid = 0, mx = 0;for (int i = 1; i <= n; ++i){if (i < mx) v[i] = min(v[mid * 2 - i], mx - i);else v[i] = 1;while(s[i + v[i]] == s[i - v[i]]) v[i]++;if (i + v[i] > mx){mx = i + v[i];mid = i;}r[i - v[i] + 1] = max(r[i - v[i] + 1], v[i] - 1);//存下以某个点为左端的最大回文串l[i + v[i] - 1] = max(l[i + v[i] - 1], v[i] - 1);}return;
}
int main()
{cin>>str;n = str.size();s[0] = s[1] = '#';for (int i = 1; i <= n; ++i){s[i * 2] = str[i - 1];s[i * 2 + 1] = '#';}n = n * 2 + 2;s[n] = 0;Manacher();for (int i = 2; i <= n; ++i){r[i] = max(r[i], r[i - 2] - 2);//减少一个长度if (l[i] && r[i]) ans = max(ans, l[i] + r[i]);}printf("%d", ans);return 0;
}