Manacher BestCoder Round #49 ($) 1002 Three Palindromes

题目传送门

 1 /*
 2     Manacher：该算法能求最长回文串，思路时依据回文半径p数组找到第一个和第三个会文串，然后暴力枚举判断是否存在中间的回文串
 3     另外，在原字符串没啥用时可以直接覆盖，省去一个数组空间，位运算 >>1 比 /2 速度快，用了程序跑快200ms左右，位运算大法好
 4 */
 5 /************************************************
 6 Author        :Running_Time
 7 Created Time  :2015-8-1 20:10:22
 8 File Name     :B.cpp
 9 *************************************************/
10
11 #include <cstdio>
12 #include <algorithm>
13 #include <iostream>
14 #include <sstream>
15 #include <cstring>
16 #include <cmath>
17 #include <string>
18 #include <vector>
19 #include <queue>
20 #include <deque>
21 #include <stack>
22 #include <list>
23 #include <map>
24 #include <set>
25 #include <bitset>
26 #include <cstdlib>
27 #include <ctime>
28 using namespace std;
29
30 typedef long long ll;
31 const int MAXN = 2e4 + 10;
32 const int INF = 0x3f3f3f3f;
33 const int MOD = 1e9 + 7;
34 char s[MAXN*2];
35 int p[MAXN*2], fi[MAXN*2], ei[MAXN*2];
36 int len;
37
38 void Manacher(void) {
39     for (int i=len; i>=0; --i)  {
40         s[i*2+2] = s[i];    s[i*2+1] = ‘#‘;
41     }
42     s[0] = ‘$‘; len = len * 2 + 2;
43     int id = 0; p[0] = 1;
44     for (int i=2; i<len; ++i)   {
45         if (id + p[id] > i) p[i] = min (p[2*id-i], id + p[id] - i);
46         else    p[i] = 1;
47         while (s[i-p[i]] == s[i+p[i]])  p[i]++;
48         if (id + p[id] < i + p[i])  id = i;
49     }
50 }
51
52 bool judge(void) {
53     int c1 = 0, c2 = 0;
54     for (int i=2; i<len-1; ++i) {
55         if (p[i] == i)  fi[++c1] = i;
56         if (i + p[i] == len)    ei[++c2] = i;
57     }
58     for (int i=1; i<=c1; ++i)   {
59         for (int j=1; j<=c2; ++j)   {
60             int l = fi[i] + p[fi[i]];
61             int r = ei[j] - p[ei[j]];
62             if (l > r)  continue;
63             int mid = (l + r) >> 1;
64             if (p[mid] > (r - l + 1) >> 1)   return true;
65         }
66     }
67     return false;
68 }
69
70 int main(void)    {       //BestCoder Round #49 ($) 1002 Three Palindromes
71     int T;  scanf ("%d", &T);
72     while (T--) {
73         scanf ("%s", s);
74         len = strlen (s);
75         if (len < 3)    {
76             puts ("NO");    continue;
77         }
78         Manacher ();
79         judge () ? puts ("Yes") : puts ("No");
80     }
81
82     return 0;
83 }