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] != '\0'; ++i)

17     {

18         s2[j++] = s1[i];

19         s2[j++] = '#';

20     }

21     s2[j] = '\0';

22 }

23 

24 void manacher(char s[])

25 {

26     int id, mx = 0;

27     p[0] = 0;

28     for(int i = 1; s[i] != '\0'; ++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] != '\0'; ++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\n", kase++, getans());

65     }

66     return 0;

67 }

代码君