POJ (Manacher) Palindrome

多敲几个模板题，加深一下对Manacher算法的理解。

这道题给的时间限制15s，是我见过的最长的时间的了。看来是为了让一些比较朴素的求最大回文子串的算法也能A过去

Manacher算法毕竟给力，运行时间200+MS

 //#define LOCAL
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;
 
 const int maxn =  + ;
 char s1[maxn], s2[maxn * ];
 int p[maxn * ];
 
 void Init(void)
 {
     s2[] = '$', s2[] = '#';
     int j = ;
     for(int i = ; s1[i] != '\0'; ++i)
     {
         s2[j++] = s1[i];
         s2[j++] = '#';
     }
     s2[j] = '\0';
 }
 
 void manacher(char s[])
 {
     int id, mx = ;
     p[] = ;
     for(int i = ; s[i] != '\0'; ++i)
     {
         if(mx > i)
             p[i] = min(p[id*-i], mx-i);
         else
             p[i] = ;
         while(s[i + p[i]] == s[i - p[i]])
             ++p[i];
         if(i + p[i] > mx)
         {
             mx = i + p[i];
             id = i;
         }
     }
 }
 
 int getans(void)
 {
     int ans = ;
     for(int i = ; s2[i] != '\0'; ++i)
         ans = max(ans, p[i] - );
     return ans;
 }
 
 int main(void)
 {
     #ifdef LOCAL
         freopen("3974in.txt", "r", stdin);
     #endif
 
     int kase = ;
     while(scanf("%s", s1) != EOF)
     {
         if(s1[] == 'E')    break;
         Init();
         manacher(s2);
         printf("Case %d: %d\n", kase++, getans());
     }
     return ;
 }

代码君