Manacher算法

Manacher

可以在(O(n))的时间内求一个字符串的最长回文子串。

教学视频链接

例题

HDU - 3068

#include <bits/stdc++.h>

#define FOPI freopen("in.txt", "r", stdin)
#define FOPO freopen("out.txt", "w", stdout)
#define FOR(i,x,y) for (int i = x; i <= y; i++)
#define ROF(i,x,y) for (int i = x; i >= y; i--)

using namespace std;
typedef long long LL;
const int inf = 0x3f3f3f3f;
const int maxn = 2e5 + 100;

char Ma[maxn*2];
int Mp[maxn*2];

void Manacher(char s[], int len)
{
    int l = 0;
    Ma[l++] = '$';
    Ma[l++] = '#';
    FOR(i, 0, len-1) Ma[l++] = s[i], Ma[l++] = '#';
    Ma[l] = 0;

    int mx = 0, id = 0;
    FOR(i, 0, l-1)
    {
        Mp[i] = mx > i ? min(Mp[2*id-i], mx-i) : 1;
        while(Ma[i+Mp[i]] == Ma[i-Mp[i]]) Mp[i]++;
        if (i+Mp[i] > mx) mx = i+Mp[i], id = i;
    }
}

char s[maxn];
int main()
{
    while(~scanf("%s", &s))
    {
        int len = strlen(s);
        Manacher(s, len);
        int ans = 0;
        FOR(i, 0, 2*len+2-1) ans = max(ans, Mp[i]-1);
        printf("%d
", ans);
    }
}