SPOJ687---repeats-repeats (suffix array +rmq)

A string S is called an (k,l)-repeat if S is obtained by concatenating k>=1 times some seed string T with length l>= 1. For example, the string

s = Abaabaabaaba

is a (4,3)-repeat with t = ABA as its seed string. That's, the seed string T is 3 characters long, and the whole string s are obtained by repeating T 4 times.

Write a program for the following Task:your program is given a long string u consisting of characters ' a ' and/or ' B ' as I Nput. Your program must find some (k,l)-repeat, occurs as substring within U with K as large as possible. For example, the input string

U = Babbabaabaabaabab

Contains the underlined (4,3)-repeat s starting at position 5. Since u contains no other contiguous substring with more than 4 repeats, your program must output the maximum K.

Input

The first line of the input contains h-the number of test cases (H <= 20). H test Cases follow. First line of all test cases is n-length of the input string (n <= 50000), the next n lines contain the input string , one character (either ' a ' or ' B ') per line, in order.
Output

For each test cases, you should write exactly one Interger K in a line-the repeat count of is maximized.
Example

Input:
1
17
B
A
B
B
A
B
A
A
B
A
A
B
A
A
B
A
B

Output:
4

Since a (4, 3)-repeat is found starting at the 5th character of the input string.

Method or the method of the paper, I use 2 suffix array, the other one is the string inversion to find out, and then respectively LCP

/************************************************************************* > File name:spoj-687.cpp > Aut Hor:alex > Mail: [email protected] > Created time:2015 April 07 Tuesday 18:16 49 seconds ******************************* *****************************************/#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <queue>#include <stack>#include <map>#include <bitset>#include <set>#include <vector>using namespace STD;Const DoublePI =ACOs(-1.0);Const intINF =0x3f3f3f3f;Const DoubleEPS =1e-15;typedef Long LongLL;typedefPair <int,int> PLL;classsuffixarray{ Public:Static Const intN =52000;intInit[n];intX[n];intY[n];intRank[n];intSa[n];intHeight[n];intBuc[n];intLog[n];intdp[n][ -];intSizevoidClear () {size =0; }voidInsertintN) {init[size++] = n; }BOOLcmpint*r,intAintBintL) {return(R[a] = = R[b] && r[a + l] = = R[b + L]); }voidGetsa (intm = the)//m is generally the maximum value of +1{Init[size] =0;intL, p, *x = x, *y = y, n = size +1; for(inti =0; I < m; ++i) {Buc[i] =0; } for(inti =0; I < n;            ++i) {++buc[x[i] = init[i]]; } for(inti =1; I < m; ++i) {Buc[i] + = buc[i-1]; } for(inti = n-1; I >=0;            -i) {sa[--buc[x[i]] = i; } for(L =1, p =1; L <= N && p < n; m = p, l *=2) {p =0; for(inti = n-l; I < n;                ++i) {y[p++] = i; } for(inti =0; I < n; ++i) {if(Sa[i] >= L)                    {y[p++] = sa[i]-l; }                } for(inti =0; I < m; ++i) {Buc[i] =0; } for(inti =0; I < n;                ++i) {++buc[x[y[i]]; } for(inti =1; I < m; ++i) {Buc[i] + = buc[i-1]; } for(inti = n-1; I >=0;                -i) {sa[--buc[x[y[i]]] = y[i]; }intI for(Swap (x, y), x[sa[0]] =0, p =1, i =1; I < n; ++i) {X[sa[i]] = cmp (y, Sa[i-1], Sa[i], L)? P-1: p++; }            }        }voidGetHeight () {inth =0, n = size; for(inti =0; I <= N;            ++i) {Rank[sa[i]] = i; } height[0] =0; for(inti =0; I < n; ++i) {if(H >0) {--h; }intJ =sa[rank[i]-1]; for(; i + H < n && j + H < n && init[i + h] = = Init[j + h]; ++h); Height[rank[i]-1] = h; }        }//preprocessing logarithm of each number for RMQ, constant optimization        voidInitlog () {log[0] = -1; for(inti =1; i < N; ++i) {Log[i] = (I & (i-1)) ? Log[i-1]: Log[i-1] +1; }        }voidINITRMQ () {initlog ();intn = size;intLimit for(inti =0; I < n; ++i) {dp[i][0] = Height[i]; } for(intj =1; J <= Log[n]; ++J) {limit = N-(1<< j); for(inti =0; I <= limit; ++i) {Dp[i][j] = min (Dp[i][j-1], Dp[i + (1<< (J-1))][j-1]); }            }        }intLCP (intAintb) {intT            A = Rank[a]; b = Rank[b];if(A > B)            {Swap (A, b);            }--b; t = log[b-a +1];returnMin (dp[a][t], dp[b-(1<< T) +1][t]); }}SA1, SA2;Charbuf[52000];Charstr[Ten];intMain () {intTscanf("%d", &t); while(t--)        {sa1.clear (); Sa2.clear ();intNscanf("%d", &n); for(inti =0; I < n; ++i) {scanf('%s ', str); Sa1.insert ((int) str[0]); Buf[i] = str[0]; } Buf[n] =' + '; for(inti = n-1; I >=0; -I.) {Sa2.insert (int) buf[i]); } Sa1.getsa ((int)(' B ') +1); Sa2.getsa ((int)(' B ') +1);        Sa1.getheight ();        Sa2.getheight ();        SA1.INITRMQ (); SA2.INITRMQ ();intAns =1, S, ans_l =0; for(intL =1; L <= N; ++L) { for(inti =0; I < n; ++i) {intA = L * i;intb = L * (i +1);if(b >= N) { Break; }intL1 = SA1. LCP (A, b);intL2 = SA2. LCP (N-a-1, N-b-1);intK = L1 + L2;if(k >0) {--k; }intCNT = k/l +1;if(cnt > ANS)                    {ans = cnt;                    s = A;                ans_l = L; }             }        }printf("%d\n", ans); }return 0;}

SPOJ687---repeats-repeats (suffix array +rmq)