HDU 5340 Three palindromes (binary enumeration +manacher+ record interval)

Three palindromes Time limit:2000/1000 MS (java/others) Memory limit:65536/65536 K (java/others)
Total submission (s): 809 Accepted Submission (s): 240


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

Inputfirst line contains a single integer T ≤ 20 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 ≤ < Span class= "Mrow" id= "mathjax-span-11" > | s | ≤ 20000

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

Sample Input

2abcabaadada

Sample Output

YesNo

Sourcebestcoder Round #49 ($)
Recommendhujie

Approximate test instructions:

Determines whether a string of length 2e4 can be divided into three palindrome strings

General idea:

BC Because of the problem, got the money 2333

Manacher processing a palindrome range, then we get two arrays, A, B, respectively: character 0 position to A[pos] is a palindrome string, character len-1 position to B[pos] is a palindrome string

Then binary enumeration A[x]~b[y] is not a palindrome string, with Manachero (1) can be sentenced to come out

Manacher calculates the longest palindrome string that has been s[x], and then pushes it back to the same interval on the original string.

Previously pushed a formula, already used this formula over three questions, it is really no problem

int L = (I-mp[i]) >>1;
int R = (i+mp[i]-4) >>1;

#pragma COMMENT (linker, "/stack:1024000000,1024000000") #include <iostream> #include <cstring> #include <cmath> #include <queue> #include <stack> #include <map> #include <set> #include < string> #include <vector> #include <cstdio> #include <ctime> #include <bitset> #include < Algorithm> #define SZ (x) ((int) (x). Size ()) #define ALL (v) (v). Begin (), (v). End () #define A foreach (I, v) for (__typeof (v ). Begin ()) I = (v). Begin (); I! = (v). End (); + + i) #define REP (i,n) for (int i=1; i<=int (n); i++) using namespace std;typedef long long ll; #define X First#define Y        Secondtypedef pair<int,int> pii;template <class t>inline bool RD (T &ret) {char c; int sgn;        if (c = GetChar (), c = = EOF) return 0;        while (c! = '-'-' && (c< ' 0 ' | | c> ' 9 ')) C = GetChar (); SGN = (c = = '-')?        -1:1; ret = (c = = '-')?        0: (C-' 0 '); while (c = GetChar (), C >= ' 0 ' &&c <= ' 9 ') ret= RET * + (C-' 0 ');        RET *= SGN; return 1;}                Template <class t>inline void PT (T x) {if (x < 0) {Putchar ('-');        x =-X;        } if (x > 9) pt (X/10); Putchar (x 10 + ' 0 ');}        const int N = 20000+100;int a[n],top1;int b[n],top2;char s[n];void ini () {top1 = 0; TOP2 = 0;}        int Mp[n<<1];char ma[n<<1];int vs[n];void manacher (char s[],int len) {int L = 0;        ma[l++] = ' $ ';        ma[l++] = ' # ';                for (int i=0;i<len;i++) {Ma[l] = s[i]; Vs[i] = l++;        Record map to position in the original string ma[l++] = ' # ';        } Ma[l] = 0;        int mx = 0,id = 0;                for (int i = 1;i < L; i + +) {mp[i]= mx>i? min (mp[2*id-i],mx-i): 1;                while (ma[mp[i]+i] = = Ma[i-mp[i]]) mp[i]++;                        if (Mp[i]+i > mx) {mx = mp[i]+i;                id = i; } if (Mp[i] >= 2){int L = (I-mp[i]) >>1; Inverse to the range left endpoint int R = (i+mp[i]-4) in the original string >>1;                        Inverse to the interval right end of the original string if (L = = 0) A[++top1] = R;                if (R = = len-1) B[++top2] = L;        }}}int Main () {int T;        RD (T);                while (t--) {scanf ("%s", s);                INI ();                int len = strlen (s);                Manacher (S,len);                Sort (A+1,A+1+TOP1);                Sort (B+1,B+1+TOP2);                bool OK = 0;                        REP (I,TOP1) {if (ok) break;                                for (int j = TOP2; J >= 1 && ok = = 0;j--) {int L = A[i], r = b[j];                                if (r-l <= 1) break;                                l++,r--;                                        if ((r-l+1) &1) {int pos = L + (r-l+1)/2; int L = (vs[Pos]-mp[vs[pos]]) >>1;                                        int R = (vs[pos]+mp[vs[pos]]-4) >>1;                                if (l <= l && R <= r) OK = 1;                                        } else {int pos = L + (r-l+1)/2;                                        int p = vs[pos]-1;                                        if (Mp[p] < 2) continue;                                        int L = (P-mp[p]) >>1;                                        int R = (p+mp[p]-4) >>1;                                if (l <= l && R <= r) OK = 1;                }}} if (OK) puts ("Yes");        Else puts ("No"); } return 0;}

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HDU 5340 Three palindromes (binary enumeration +manacher+ record interval)