Hotaru's problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 907 Accepted Submission(s): 322
Problem Description
Hotaru Ichijou recently is addicated to math problems. Now she is playing with N-sequence.
Let's define N-sequence, which is composed with three parts and satisfied with the following condition:
1. the first part is the same as the thrid part,
2. the first part and the second part are symmetrical.
for example, the sequence 2,3,4,4,3,2,2,3,4 is a N-sequence, which the first part 2,3,4 is the same as the thrid part 2,3,4, the first part 2,3,4 and the second part 4,3,2 are symmetrical.
Give you n positive intergers, your task is to find the largest continuous sub-sequence, which is N-sequence.
Let's define N-sequence, which is composed with three parts and satisfied with the following condition:
1. the first part is the same as the thrid part,
2. the first part and the second part are symmetrical.
for example, the sequence 2,3,4,4,3,2,2,3,4 is a N-sequence, which the first part 2,3,4 is the same as the thrid part 2,3,4, the first part 2,3,4 and the second part 4,3,2 are symmetrical.
Give you n positive intergers, your task is to find the largest continuous sub-sequence, which is N-sequence.
Input
There are multiple test cases. The first line of input contains an integer T（T<=20）, indicating the number of test cases.
For each test case:
the first line of input contains a positive integer N（1<=N<=100000）, the length of a given sequence
the second line includes N non-negative integers ,each interger is no larger than 109 , descripting a sequence.
For each test case:
the first line of input contains a positive integer N（1<=N<=100000）, the length of a given sequence
the second line includes N non-negative integers ,each interger is no larger than 109 , descripting a sequence.
Output
Each case contains only one line. Each line should start with “Case #i: ”,with i implying the case number, followed by a integer, the largest length of N-sequence.
We guarantee that the sum of all answers is less than 800000.
We guarantee that the sum of all answers is less than 800000.
Sample Input
1
10
2 3 4 4 3 2 2 3 4 4
Sample Output
Case #1: 9
Source
解题：manacher暴力搞,fread真是TLE的救星啊
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 201000; 4 int s[maxn],p[maxn]; 5 int manacher(int n) { 6 int id = 0,maxlen = 0; 7 s[0] = -1; 8 for(int i = n; i >= 0; --i) { 9 s[i + i + 2] = s[i]; 10 s[i + i + 1] = 0; 11 } 12 n = (n<<1|1); 13 for(int i = 2; i < n; ++i) { 14 if(p[id] + id > i) p[i] = min(p[2*id-i],p[id]+id-i); 15 else p[i] = 1; 16 while(s[i-p[i]] == s[i+p[i]]) ++p[i]; 17 if(id + p[id] < i + p[i]) id = i; 18 if(maxlen < p[i]) maxlen = p[i]; 19 } 20 return maxlen - 1; 21 } 22 char *ch, *ch1, buf[50*1024000+5], buf1[50*1024000+5]; 23 void read(int &x) { 24 for (++ch; *ch <= 32; ++ch); 25 for (x = 0; '0' <= *ch; ch++) x = x * 10 + *ch - '0'; 26 } 27 int main() { 28 int kase,n,cs = 1; 29 ch = buf - 1; 30 ch1 = buf1 - 1; 31 fread(buf, 1, 1000 * 45 * 1024, stdin); 32 read(kase); 33 while(kase--) { 34 read(n); 35 for(int i = 0; i < n; ++i) 36 read(s[i]); 37 manacher(n); 38 int ret = 0; 39 n = (n<<1|1); 40 for(int i = 3; i < n; i += 2) { 41 if(p[i] - 1 > ret) { 42 int r = p[i] + 1; 43 while(r > ret && p[i + r] < r) --r; 44 if(ret < r) ret = r; 45 } 46 } 47 printf("Case #%d: %d ",cs++,(ret>>1)*3); 48 } 49 return 0; 50 } 51 /* 52 1 53 11 54 1 1 2 1 1 2 1 1 2 1 1 55 */