题目传送门
1 /*
2 题意：求形如(2 3 4) (4 3 2) (2 3 4)的最长长度，即两个重叠一半的回文串
3 Manacher：比赛看到这题还以为套个模板就行了，因为BC上有道类似的题，自己又学过Manacher算法，结果入坑WA到死
4 开始写的是判断是否 p[i]-1 <= p[i+p[i]-1]-1，但是没有想到这种情况：5 (5 1) (1 5) (5 1) 1
5 单靠最长回文半径是不行的，看了网上的解题报告知道，要从极端位置往回挪才行
6 给我的教训是只会套模板是没用的，要灵活的使用该算法。另外max() 函数似乎很费时间
7 详细解释
8 */
9 /************************************************
10 * Author :Running_Time
11 * Created Time :2015-8-11 12:52:49
12 * File Name :C.cpp
13 ************************************************/
14
15 #include <cstdio>
16 #include <algorithm>
17 #include <iostream>
18 #include <sstream>
19 #include <cstring>
20 #include <cmath>
21 #include <string>
22 #include <vector>
23 #include <queue>
24 #include <deque>
25 #include <stack>
26 #include <list>
27 #include <map>
28 #include <set>
29 #include <bitset>
30 #include <cstdlib>
31 #include <ctime>
32 using namespace std;
33
34 #define lson l, mid, rt << 1
35 #define rson mid + 1, r, rt << 1 | 1
36 typedef long long ll;
37 const int MAXN = 1e5 + 10;
38 const int INF = 0x3f3f3f3f;
39 const ll INFF = 0x7fffffff;
40 const int MOD = 1e9 + 7;
41 int a[MAXN*2], p[MAXN*2];
42 int n;
43
44 int Manacher(void) {
45 a[n] = INF + 2;
46 for (int i=n; i>=0; --i) {
47 a[i*2+2] = a[i]; a[i*2+1] = INF;
48 }
49 a[0] = INF + 1; n = n * 2 + 2;
50 int id = 0; p[0] = 1;
51 for (int i=2; i<n; ++i) {
52 if (id + p[id] > i) p[i] = min (p[2*id-i], id + p[id] - i);
53 else p[i] = 1;
54 while (a[i-p[i]] == a[i+p[i]]) p[i]++;
55 if (id + p[id] < i + p[i]) id = i;
56 }
57
58 int mx = 0;
59 for (int i=3; i<=n-2; i+=2) {
60 if (p[i] - 1 > mx) {
61 int c = p[i] - 1;
62 while (c > mx && p[i+c] < c) c--;
63 mx = mx > c ? mx : c;
64 }
65 }
66
67 return mx / 2 * 3;
68 }
69
70 int main(void) { //HDOJ 5371 Hotaru's problem
71 int T, cas = 0; scanf ("%d", &T);
72 while (T--) {
73 scanf ("%d", &n);
74 for (int i=0; i<n; ++i) scanf ("%d", &a[i]);
75 printf ("Case #%d: %d
", ++cas, Manacher ());
76 }
77
78 return 0;
79 }