UESTC Search Topic m topic palindromic String Horse-drawn vehicle algorithm

Palindromic String

Time limit:20 Sec Memory limit:256 MB

Topic Connection HTTP://ACM.UESTC.EDU.CN/#/CONTEST/SHOW/61
Description

Qiushi Big Brother likes to explore new things, recently he invented a new palindrome string, called K-heavy palindrome string! Today he wants to use it to test the children.

Qiushi the information about the K-heavy palindrome string.

Any string belongs to a 0 palindrome string, including an empty string.
A string of length n s,s is a K (k≥1) re-palindrome string, when and only if S is a palindrome string, and its length is ⌊n2⌋ prefix and the length of the ⌊n2⌋ suffix is k−1 heavy palindrome string.
If a string is a K-palindrome string, the string is said to have a palindrome value of K. A string can have more than one palindrome value, such as S=abaaba, its palindrome value can be 0,1,2,3.
The maximum palindrome value of a string is the maximum value of all palindrome values for that string.
If the maximum palindrome value of the string s ≥1, then s must be a palindrome string.
A string s, if it is the same as read and reverse read, it is called S is palindrome string, such as Aabaa,aba,a. But ABC,ABAB,AACBA is not a palindrome.
A string of length n s, which has n+1 prefixes and n+1 suffixes (not necessarily non-null), such as ABCDE, has 6 prefixes, respectively, is an empty string, A,AB,ABC,ABCD,ABCDE, there are 6 suffixes, respectively, is an empty string, E,DE,CDE,BCDE,ABCDE.

Qiushi Big Brother gives you a string s, he wants to ask you, s the maximum palindrome for all prefixesa is completely coincident, or can be shifted up or down by a line < Span id= "mathjax-span-1329" class= "math" style= "WIDTH:0.958EM; Display:inline-block; " >a a section of the complete coincidence, it means that Qiushi Big Brother blew out a part of the melody of her sister. Input first line Enter a string s (0<| s|≤2⋅106), S contains uppercase English letter (A-Z), lowercase English letter (A-Z), number (0-9) Output

Outputs an integer that represents the sum of the maximum palindrome values for all prefixes in S .

Sample InputCcececccee
Sample Output4hint

Test Instructions

Exercises

First run the horse-drawn cart algorithm, and then get all the characters of the P array, indicating the length of the palindrome string
And then we'll solve it recursively.
And then you know that each palindrome is a few heavy
(although I think the complexity may be very high, but run quickly, do not know why ...)

Code:

//Qscqesze#include <cstdio>#include<cmath>#include<cstring>#include<ctime>#include<iostream>#include<algorithm>#include<Set>#include<vector>#include<sstream>#include<queue>#include<typeinfo>#include<fstream>#include<map>#include<stack>typedefLong Longll;using namespacestd;//freopen ("d.in", "R", stdin);//freopen ("D.out", "w", stdout);#defineSspeed ios_base::sync_with_stdio (0); Cin.tie (0)#defineMAXN 2000010#defineMoD 10007#defineEPS 1e-9intNum;Charch[ -];//const int INF=0X7FFFFFFF; //нчоч╢сConst intinf=0x3f3f3f3f;/*inline void P (int x) {num=0;if (!x) {Putchar (' 0 ');p UTS (""); return;}    while (x>0) ch[++num]=x%10,x/=10;    while (Num) Putchar (ch[num--]+48); Puts ("");}*/inline ll read () {intx=0, f=1;CharCh=GetChar ();  while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();}  while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; ch=GetChar ();} returnx*F;} InlinevoidPintx) {Num=0;if(!x) {Putchar ('0');p UTS ("");return;}  while(x>0) ch[++num]=x%Ten, x/=Ten;  while(Num) Putchar (ch[num--]+ -); Puts ("");}//**************************************************************************************CharS[MAXN];Charstr[maxn*2];intl;intp[maxn*2];intManacher (CharS[],intl) {   inti,j,k,ans=0;  for(i=1; i<=l;++i) str[i<<1]=s[i],str[(i<<1)+1]='#'; str[1]='#'; str[l*2+1]='#'; str[0]='&'; str[l*2+2]='$'; L=l*2+1; j=0;  for(i=1; i<=l;) {        while(str[i-j-1]==str[i+j+1])++J; P[i]=j;if(J>ans) ans=J;  for(k=1; k<=j&&p[i]-k!=p[i-k];++k) P[i+k]=min (p[i-k],p[i]-k); I+=k;j=max (J-k,0); }   returnans;}intMain () {scanf ("%s", s+1); L=strlen (s+1);    Manacher (s,l); L=l*2+1; intans=0;  for(intI=2; i<= (L +1)/2; i++)    {        inttmp=P[i]; intpos=i;  while(tmp+1>=pos&&tmp!=0) {ans++; POS= (pos+1)/2; TMP=P[pos]; }        //cout<<ans<<endl;} cout<<ans<<Endl;}

UESTC Search Topic m topic palindromic String Horse-drawn vehicle algorithm