poj3974 Palindrome(Manacher最长回文)

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 const int maxn = 1e6 + 5;
 8 char s[maxn], tmp[2*maxn];
 9 int p[maxn*2], id, mx, ans;
10 inline void manacher(){
11     mx = id = ans = 0;
12     memset( p, 0, sizeof(p) );
13     for( int i=1; tmp[i]; i++ ){
14         p[i] = mx>i ? min(p[2*id-i], mx-i):1;
15         while( tmp[i+p[i]]==tmp[i-p[i]] ) p[i]++;
16         if( mx<i+p[i] ){
17             mx = i+p[i];
18             id = i;
19         }
20     }
21     for( int i=1; tmp[i]; i++ )
22         if( ans<p[i] ) ans = p[i];
23 }
24 
25 int main(){
26     int kase = 0;
27     while( ~scanf("%s", s+1) && s[1]!=‘E‘ ){
28         int len = strlen(s+1);
29         tmp[0] = ‘$‘;
30         int j = 2;
31         for( int i=1; i<=len; i++, j+=2 ){
32             tmp[j+1] = tmp[j-1] = ‘#‘;
33             tmp[j] = s[i];
34         }
35         tmp[j+2] = 0;
36         manacher();
37         printf("Case %d: %d\n", ++kase, ans-1);
38     }
39 
40     return 0;
41 }
42 /*
43 Sample Input
44 
45 abcbabcbabcba
46 abacacbaaaab
47 END
48 Sample Output
49 
50 Case 1: 13
51 Case 2: 6
52 */