poj 3974 Palindrome (最长回文字串，manacher裸题)

poj3974 题意：求最大长度的回文字串。思路：manacher裸题，用来练习算法。
代码实现
 1/* ***********************************************
 2Author :111qqz
 3Created Time :2016年04月18日 星期一 16时32分25秒
 4File Name :code/poj/3974.cpp
 5************************************************ */
 6
 7#include <cstdio>
 8#include <cstring>
 9#include <iostream>
10#include <algorithm>
11#include <vector>
12#include <queue>
13#include <set>
14#include <map>
15#include <string>
16#include <cmath>
17#include <cstdlib>
18#include <ctime>
19#define fst first
20#define sec second
21#define lson l,m,rt<<1
22#define rson m+1,r,rt<<1|1
23#define ms(a,x) memset(a,x,sizeof(a))
24typedef long long LL;
25#define pi pair < int ,int >
26#define MP make_pair
27
28using namespace std;
29const double eps = 1E-8;
30const int dx4[4]={1,0,0,-1};
31const int dy4[4]={0,-1,1,0};
32const int inf = 0x3f3f3f3f;
33const int N=2E6+7;
34char st[2*N];
35int p[2*N];
36
37int manacher(char *s)
38{
39    int len = strlen(s);
40    for ( int i = len ; i >= 0 ; i-- )
41    {
42	s[i+i+2]=s[i];
43	s[i+i+1]='#';
44    }
45    s[0]='$';
46
47    int id = 0 ;
48    int maxlen = 0 ;
49
50    for ( int i = 2 ; i < 2*len+1 ; i++)
51    {
52	if (id+p[id]>i) p[i] = min(p[2*id-i],id+p[id]-i);
53	else p[i] = 1;
54
55	while (s[i+p[i]]==s[i-p[i]]) p[i]++;
56	if (id+p[id]<i+p[i]) id =  i;
57	if (p[i]>maxlen) maxlen = p[i];
58    }
59    return maxlen-1;
60}
61int main()
62{
63	#ifndef  ONLINE_JUDGE
64	freopen("code/in.txt","r",stdin);
65  #endif
66	int cas = 0 ;
67	while (scanf("%s",st)!=EOF)
68	{
69	    if (st[0]=='E'&&st[1]=='N'&&st[2]=='D') break;
70	    printf("Case %d: %d\n",++cas,manacher(st));
71
72	}
73
74  #ifndef ONLINE_JUDGE
75  fclose(stdin);
76  #endif
77    return 0;
78}
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