POJ 3261 Milk Patterns suffix array to find a string of the longest repeatable substring repeat at least k times

Milk Patterns

Description

Farmer John has noticed, the quality of milk given by he cows varies from day to day. On further investigation, he discovered so although he can ' t predict the quality of milk from one day to the next, there is some regular patterns in the daily milk quality.

To perform a rigorous study, he had invented a complex classification scheme by which each milk sample was recorded as an I Nteger between 0 and 1,000,000 inclusive, and have recorded data from a single cow over n (1≤n ≤20,000) Days. He wishes to find the longest pattern of samples which repeats identically at least K (2≤ k ≤ N) times. This could include overlapping patterns--1 2 3 2 3 2 3 1 repeats 2 3 2 3 twice, for example.

Help Farmer John by finding the longest repeating subsequence in the sequence of samples. It is guaranteed, at least one subsequence are repeated at least K times.

Input

Line 1:two space-separated integers: Nand K
Lines 2.. N+ 1: NIntegers, one per line, the quality of the milk on day IAppears on the ITh line.

Output

Line 1:one Integer, the length of the longest pattern which occurs at least KTimes

Sample Input

8 212323231

Sample Output

4

Remember the discretization of

#include <iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#include<vector>using namespacestd;#pragmaComment (linker, "/stack:102400000,102400000")#defineLS i<<1#defineRS ls | 1#defineMid ((LL+RR) >>1)#definePII pair<int,int>#defineMP Make_pairtypedefLong LongLL;Const Long LongINF = 1e18+1LL;Const DoublePi = ACOs (-1.0);Const intN = 2e5+Ten, M = 2e5+ -, mod = 1e9+7, INF =2e9;///Heght[i] represents the longest common prefix for Suffix (sa[i-1]) and Suffix (Sa[i]):///Rank[i] Indicates the rank of the suffix that begins with I:///Sa[i] Indicates the beginning of the suffix that is ranked I:int*Rank,r[n],sa[n],height[n],wa[n],wb[n],wm[n];BOOLcmpint(RNintAintBintl) {returnR[a] = = R[b] && r[a+l] = = r[b+l];}voidSA (int(RNint*sa,intNintm) {int*x=wa,*y=wb,*T;  for(intI=0; i<m;++i) wm[i]=0;  for(intI=0; i<n;++i) wm[x[i]=r[i]]++;  for(intI=1; i<m;++i) wm[i]+=wm[i-1];  for(inti=n-1; i>=0;-I.) sa[--wm[x[i]]]=i;  for(intI=0, j=1, p=0;p <n;j=j*2, m=p) {         for(p=0, I=n-j;i<n;++i) y[p++]=i;  for(i=0; i<n;++i)if(SA[I]&GT;=J) y[p++]=sa[i]-J;  for(i=0; i<m;++i) wm[i]=0;  for(i=0; i<n;++i) wm[x[y[i]]]++;  for(i=1; i<m;++i) wm[i]+=wm[i-1];  for(i=n-1; i>=0;-I.) sa[--wm[x[y[i]]]]=Y[i];  for(t=x,x=y,y=t,i=p=1, x[sa[0]]=0; i<n;++i) {X[sa[i]]=CMP (y,sa[i],sa[i-1],j)? p1:p + +; }} Rank=x;}voidHeight (int(RNint*sa,intN) { for(intI=0, j=0, k=0; i<n;height[rank[i++]]=k) for(k?--K:0, j=sa[rank[i]-1];r[i+k] = = r[j+k];++k);}intn,a[n],k;intCheckintx) {inti =2, CNT;  while(1) {             while(I <= n && height[i] < x) i++; if(i > N) Break; CNT=1;  while(I <= n && height[i] >=x) {cnt++; I++; }            if(CNT >=k) {return 1; }        }        return 0;}intMain () { while(~SCANF ("%d%d",&n,&k)) { for(inti =1; I <= N; ++i) scanf ("%d", &a[i]), r[i-1] =A[i]; Sort (a+1, a+n+1); intc = Unique (A +1, a+n+1)-A-1;  for(inti =0; I < n; ++i) R[i] = Lower_bound (A +1, a+c+1, R[i])-A; R[n]=0; SA (R,sa,n+1,20001);            Height (R,sa,n); intLL =1, rr = N,ans =0;  while(LL <=RR) {                intMD = (ll + RR) >>1; if(Check (MD)) {ans= MD, LL = MD +1; } ElseRR = MD-1; } printf ("%d\n", ans); }        return 0;}

　　

POJ 3261 Milk Patterns suffix array to find a string of the longest repeatable substring repeat at least k times