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

 1 //#define LOCAL
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <algorithm>
 6 using namespace std;
 7 
 8 const int maxn = 1000000 + 100;
 9 char s1[maxn], s2[maxn * 2];
10 int p[maxn * 2];
11 
12 void Init(void)
13 {
14     s2[0] = '$', s2[1] = '#';
15     int j = 2;
16     for(int i = 0; s1[i] != ''; ++i)
17     {
18         s2[j++] = s1[i];
19         s2[j++] = '#';
20     }
21     s2[j] = '';
22 }
23 
24 void manacher(char s[])
25 {
26     int id, mx = 0;
27     p[0] = 0;
28     for(int i = 1; s[i] != ''; ++i)
29     {
30         if(mx > i)
31             p[i] = min(p[id*2-i], mx-i);
32         else
33             p[i] = 1;
34         while(s[i + p[i]] == s[i - p[i]])
35             ++p[i];
36         if(i + p[i] > mx)
37         {
38             mx = i + p[i];
39             id = i;
40         }
41     }
42 }
43 
44 int getans(void)
45 {
46     int ans = 1;
47     for(int i = 1; s2[i] != ''; ++i)
48         ans = max(ans, p[i] - 1);
49     return ans;
50 }
51 
52 int main(void)
53 {
54     #ifdef LOCAL
55         freopen("3974in.txt", "r", stdin);
56     #endif
57 
58     int kase = 1;
59     while(scanf("%s", s1) != EOF)
60     {
61         if(s1[0] == 'E')    break;
62         Init();
63         manacher(s2);
64         printf("Case %d: %d
", kase++, getans());
65     }
66     return 0;
67 }

代码君