Manacher HDOJ 5371 Hotaru's problem

 

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\n", ++cas, Manacher ());
76     }
77 
78     return 0;
79 }

 

转载于:https://www.cnblogs.com/Running-Time/p/4723498.html