【BZOJ 2565】 最长双回文串

【题目链接】

https://www.lydsy.com/JudgeOnline/problem.php?id=2565

【算法】

Manacher

【代码】

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + ;

char s[MAXN];

inline void Manacher()
{
        int i,pos = ,mx = ,len,ans = ;
        static char tmp[MAXN<<];
        static int p[MAXN<<],l[MAXN<<],r[MAXN<<];
        len = strlen(s+);
        for (i = ; i <= len; i++)
        {
                tmp[*i-] = '#';
                tmp[*i] = s[i];
        }
        tmp[len =  * len + ] = '#';
        for (i = ; i <= len; i++)
        {
                if (mx > i) p[i] = min(p[*pos-i],mx-i);
                else p[i] = ;
                while (i - p[i] >=  && i + p[i] <= len && tmp[i-p[i]] == tmp[i+p[i]]) p[i]++;
                if (i + p[i] -  > mx)
                {
                        mx = i + p[i] - ;
                        pos = i;
                }
        }
        pos = ;
        for (i = ; i <= len; i++)
        {
                if (tmp[i] == '#')
                {
                        while (pos + p[pos] < i) pos++;
                        l[i] = i - pos;
                }
        }
        pos = len;
        for (i = len; i >= ; i--)
        {
                if (tmp[i] == '#')
                {
                        while (pos - p[pos] > i) pos--;
                        r[i] = pos - i;
                }
        }
        for (i = ; i <= len; i++) ans = max(ans,l[i]+r[i]);
        printf("%d\n",ans);
}

int main()
{

        scanf("%s",s+);
        Manacher();

        return ;

}