Hdu 5340 the longest palindrome strand deformation

http://acm.hdu.edu.cn/showproblem.php?pid=5340

Problem Descriptioncan We divided a given string S into three nonempty palindromes?

Inputfirst line contains a single integer T≤ which denotes the number of test cases.

For each test case, the there is a single line contains a string S which only consist of lowercase Chinese letters. 1≤| s | ≤ 20000

Outputfor each case with the output the "Yes" or "No" in a.

Sample Input

2abcabaadada

Sample Output

YesNo

/**hdu 5340 longest palindrome strand deformation topic: Given a string question whether it can be divided into three palindrome (not empty string) thinking: If the direct enumeration judgment Time complexity O (n^3). The Manacher algorithm can be used to find all 0~i for palindrome string and j~n-1 as palindrome string I and J last enumeration of the two positions using the original r[i], the middle is not a palindrome. No more complexity than O (n^2) */#include <iostream> #include <cmath> #include <string.h> #include <stdio.h># include<algorithm>using namespace Std;const int maxn=110010;/*** find the longest palindrome substring O (n) * abaaba* i:0 1 2 3 4 5 6 7 8 9 10 1 1 13* a[i]: $ # a # b # a # a # b # a #* r[i]: 1 1 2 1 4 1 2 7 2 1 4 1 2 1* answer= (max (r[i]-1)) */char a[maxn*2],    S[maxn];int r[maxn*2];int C1[20011],c2[20011];///0~i satisfies all i;j~n-1 of palindrome to satisfy palindrome all jvoid manacher (char s[],int n) {int l=0;    A[l++]= ' $ ';    a[l++]= ' # ';        for (int i=0; i<n; i++) {a[l++]=s[i];    a[l++]= ' # ';    } a[l]=0;    int mx=0,id=0;        for (int i=0; i<l; i++) {r[i]=mx>i?min (r[2*id-i],mx-i): 1;        while (A[i+r[i]]==a[i-r[i]]) r[i]++;            if (I+R[I]&GT;MX) {mx=i+r[i];        Id=i;  }}}int Main () {int T;  scanf ("%d", &t);        while (t--) {scanf ("%s", s);        int n=strn (s);        Manacher (S,n);        int l1=0,l2=0;                for (int i=1; i<2*n+2; i++) {if (r[i]==i && i!=1) {l1++;            C1[l1]=r[i]-1;                } if (R[i]==2*n+2-i && i!=2*n+1) {l2++;            C2[l2]=r[i]-1;        }} int flag=1;                for (int i=l1, i>=1&&flag; i--) {for (int j=1; j<=l2&&flag; J + +) {                int ll=c1[i]*2+2;                int rr=2*n+2-(c2[j]*2+2);                int tmp= (LL+RR)/2;                if (LL&GT;RR) continue;                if (r[tmp]-1==0) continue;                if (r[tmp]>= (rr-ll+2)/2) {flag=0;        }}} if (flag) puts ("No");    Else puts ("Yes"); } return 0;}

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Hdu 5340 the longest palindrome strand deformation