HDU 5340-three palindromes (manacher algorithm)

Title Address: HDU 5340
Test instructions: Ask if the string str can be decomposed into three non-empty palindrome strings.
Idea: We can get two arrays first and last by processing the string according to the Manacher algorithm and the p array produced during processing.
First stores the value of the radius possible for a palindrome string.
Last store is the radius possible value of the third palindrome string.
We can enumerate the first palindrome string and the third palindrome string, and then find the initial position and end position of the second palindrome according to the radius. Then we calculate the midpoint of the second palindrome string:
1. If LL>RR, the second string is a negative length, pass
2. If p[mid]=1, the second palindrome string is "#", equivalent to NULL, pass
3. If the length of the three substrings is greater than Len, the length of the match is proved to be the same.

#include <stdio.h>#include <math.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <sstream>#include <algorithm>#include <set>#include <queue>#include <stack>#include <map>#pragma COMMENT (linker, "/stack:102400000,102400000")using namespace STD;typedef Long LongLL;Const intinf=0x3f3f3f3f;Const DoublePi=ACOs(-1.0);Const Doubleesp=1e-6;Const intmaxn=110010;CharSTR[MAXN];Chars[maxn*2];intp[maxn*2];intLenintfirst[maxn*2];intlast[maxn*2];voidManacher () {intI s[0]=' $ '; s[1]=' # '; for(i=0; i<len;i++) {s[i*2+2]=str[i]; s[i*2+3]=' # '; } s[i*2+2]=' + '; len=2*i+2;intMaxl,id; Id=maxl=0;memset(P,0,sizeof(p)); for(intI=0; i<len; i++) {if(p[id]+id>i) P[i]=min (p[2*ID-I],P[ID]+ID-I);Elsep[i]=1; while(S[i+p[i]]==s[i-p[i]]) p[i]++;if(P[i]+i>p[id]+id)        {id=i; }if(P[I]&GT;MAXL)    Maxl=p[i]; }}intMain () {intTscanf("%d", &t); while(t--) {scanf('%s ', &AMP;STR); len=strlen(str); Manacher ();intl1=0, r1=0; for(intI=2; i<=len-2; i++) {if(I==p[i]) first[l1++]=i;if(p[i]==len-i) last[r1++]=i; }intflag=0;intLl,rr,mid; for(intI=0; i<l1;i++) { for(intj=0; j<r1;j++) {ll=first[i]+p[first[i]]; RR=LAST[J]-P[LAST[J]];if(LL&GT;RR)Continue; Mid= (LL+RR)/2;if(ll==rr&&s[mid]==' # ')Continue;if(p[first[i]]*2-1+p[mid]*2-1+p[last[j]]*2-1>=len+1) {flag=1; Break; }            }if(flag) Break; }if(flag)puts("Yes");Else            puts("No"); }return 0;}

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HDU 5340-three palindromes (manacher algorithm)