Palinwords (palindromic sequence processing)

Topic links: https://vjudge.net/contest/344930#problem/G

 

Title effect: you string, if he 3 palindrome includes at least equal to a length greater than two, and these can not be nested palindromic (e.g. nested aaa aaaa, waw nested awawa), if the string so cattle force, then it is his output.

 

Topic ideas: This question actually has a greedy thoughts in it, in fact, every time we just need to find the length of 3 and 4 string length palindromes just fine. Here is the first process of processing strategies palindromic sequence length of 3, and then the processing sequence of length 4 palindromic. Processing palindromic sequence of length 4, when it is to be avoided substring is a palindromic sequence.

 

 1 #include <stdio.h>
 2 #include <algorithm>
 3 #include <iostream>
 4 #include <stdlib.h>
 5 #include <string>
 6 #include <string.h>
 7 #include <math.h>
 8 #include <vector>
 9 #include <queue>
10 #include <stack>
11 #include <map>
12 #include <set>
13 
14 
15 #define INF 0x3f3f3f3f
16 #define LL long long
17 
18typedef unsigned long  long eye;
19  const  int maxn 1E5 + = 10 ;
20  
21  char s [maxn];
22 out base = 131 ;
23 out mod = 1e9 + 7 ;
24  out p [maxn];
25  out h1 [maxn] H2 [maxn];
26  eye q [maxn];
27  
28  
29 eye get_hash (h eye [], int s, int r) {
 30      return (h [r] - h [L- 1 ] * p [l + r- 1 ]);
31 }
32 
33 bool check(char s[]) {
34     std::set<ull> st;
35     int ans = 0;
36     int len = strlen(s+1);
37     for (int i=2;i+1<=len;i++) {
38         if (s[i-1] == s[i+1]) {
39             ull temp = get_hash(h1,i-1,i+1);
40             if (st.find(temp) == st.end()) {
41                 st.insert(temp);
42                 ans++;
43             }
44         }
45     }
46 
47     for (int i=1;i+3<=len;i++) {
48         if (s[i] == s[i+3] && s[i+1] == s[i+2]) {
49             ull a = get_hash(h1,i,i+3);
50             ull b = get_hash(h1,i+1,i+3);
51             ull c = get_hash(h1,i,i+2);
52             if (st.find(a) == st.end() && st.find(b) == st.end() && st.find(c) == st.end()) {
53                 ans++;
54                 st.insert(a);
55                 st.insert(b);
56                 st.insert(c);
57             }
58         }
59     }
60     return ans >= 2;
61 }
62 
63 int main() {
64     p[0] = 1;
65     for (int i=1;i<maxn;i++) {
66         p[i] = p[i-1] * base;
67     }
68     while (~scanf("%s",s+1)) {
69         int len = strlen(s+1);
70         for (int i=1;i<=len;i++) {
71             h1[i] = h1[i-1] * base + s[i];
72         }
73         if (check(s)) {
74             printf("%s\n",s+1);
75         }
76     }
77     return 0;
78 }