HDU 5340 three palindromes (string processing + palindrome)

HDU 5340 Three Palindromes

Problem Description
Determines whether the string s can be divided into three non-empty palindrome strings.
Enter a description
The first line is an integer t that represents the number of data groups. T \leq 20t≤20
For each group, only a string consisting of lowercase letters is included. 1 \leq | s| \leq 200001≤∣s∣≤20000
Output Description
For each group, the single-line output is "Yes" or "No".
Input Sample
2
Abc
Abaadada
Output Sample
Yes
No

Topic: Give a string, determine whether it can be divided into three non-empty palindrome string. Problem-solving idea: First use manacher algorithm to find the longest palindrome range that can be expanded at the center of each point. You can then try out two arrays of pre[i] and Pos[i], recording the cut at I point, can make the substring before or after I, become a palindrome string. After the pre and POS arrays are found, traverse the center of the middle section of the string, each finding a center point that expands to its center (1 < A < P[i]-1) when pre[i-a] and Pos[i + A] are marked, then the cut is valid.

#include <cstdio>#include <cstring>#include <algorithm>using namespace STD;Const intN =20005;Charstr[n<<1], s[n];intp[n<<1], Len, len2;voidManacher () {len =strlen(s); for(inti =0; i < Len; i++) {str[i<<1] =' # '; str[(i<<1) +1] = S[i]; } str[len<<1] =' # '; str[(len<<1) +1] =' + '; Len =strlen(str);intMAXP =0, id =0; for(inti =0; i < Len; i++) {if(Maxp > i) p[i] = min (maxp-i, p[(id<<1)-i]);ElseP[i] =1; for(; str[i-p[i] [= Str[i + p[i]] && i + p[i] < len && I-p[i] >=0; p[i]++);if(i + p[i] > Maxp)            {MAXP = p[i] + i;        id = i; }    }}BOOLa[40005], b[40005];BOOLMark () {len =strlen(s); Len2 = (len<<1)  +1;memsetA0, len2);memsetB0, len2);if(len = =3)return true;if(Len <3)return false; for(inti =1; i < len2; i++) {if(P[i]-1= = i) A[i + (P[i]-1)] =true;if(P[len2-1-I]-1= = i) B[len2-1-I-(P[LEN2-1-I]-1)] =true; } for(inti =1; I < len2-1; i++) { for(intj =1; J < P[i]; J + +) {if(A[i-j] & B[i + j]) {return true; }           }    }return false;}intMain () {intTscanf("%d", &t); while(t--) {scanf('%s ', s); Manacher ();if(Mark ()) {printf("yes\n"); }Else printf("no\n"); }return 0;}

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HDU 5340 three palindromes (string processing + palindrome)