manacher(最长回文串)

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https://www.felix021.com/blog/read.php?2040
这个网址上写的很详细,很容易看懂

#include<bits/stdc++.h>
using namespace std;

int p[1005];
string change(string s, int l)//改变字符串,变成奇数长度的串
{
    string cs = "";
    for (int i = 0; i<l; i++)
    {
        cs = cs+'#';
        cs = cs + s[i];
    }
    cs = '@' + cs + '#';
    return cs;
}
void manacher(string s, int l)
{
    int mr = 0, mm = 0;
    for (int i = 1; i<l; i++)
    {
        p[i] = mr>i ? min(p[2 * mm - i], mr - i) : 1;//如果maxright>i那么以i为中心的回文串包括在maxmid里面,并且2*mm-i这个点,是i关于maxmid对称的点,p[i]=p[2*mm-1];
        while (s[i - p[i]] == s[i + p[i]])
            p[i]++;
        if (i + p[i]>mr)//更新maxright
        {
            mr = i + p[i];
            mm = i;
        }
    }
}
int main()
{
    string s, cs;
    int l;
    cin >> cs;
    l = cs.size();
    s = change(cs, l);
    l = s.size();
    manacher(s, l);
    int ans = 0;
    for (int i = 1; i<l; i++)
    {
        ans = max(ans, p[i]-1);
    }
    cout << ans << endl;
    return 0;
}

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转载自blog.csdn.net/dpdpd/article/details/52078245