字符串回文算法模板

manacher:

const int N = 200010;
char ori[N], pat[N*2];
int p[N*2];
int manacher(char *ori)
{
    int len = strlen(ori + 1);
    for(int i = 1; i <= len; i++)
        pat[i*2] = ori[i], pat[i*2+1] = '#';
    pat[0] = '?', pat[1] = '#', pat[len*2+2] = '\0';
    int k = 0, extk = 0, ans = 0;
    for(int i = 2; pat[i]; i++)
    {
        p[i] = i < extk ? min(extk - i, p[2*k-i]) : 1;
        while(pat[i-p[i]] == pat[i+p[i]]) p[i]++;
        if(p[i] + i > extk)
            k = i, extk = i + p[i], ans = max(ans, p[i]);
    }
    return ans - 1;
}
int main()
{
    while(~ scanf(" %s", ori + 1))
        printf("%d\n", manacher(ori));
    return 0;
}

迷之回文串算法：

#include 
using namespace std;

const int N = 1000000 + 10;
char ori[N];
int fast(char *ori)
{
    int ans = 0;
    ori[0] = '?';
    for(int i = 0; ori[i]; i++)
    {
        int s = i, e = i;
        while(ori[e+1] == ori[i]) ++e;
        i = e;
        while(ori[s-1] == ori[e+1]) --s, ++e;
        ans = max(ans, e - s + 1);
    }
    return ans;
}
int main()
{
    int n;
    scanf("%d", &n);
    while(n--)
    {
        scanf("%s", ori + 1);
        printf("%d\n", fast(ori));
    }
    return 0;
}