Hihocoder #1407: suffix array two • Repeat Melody 2

#1407: suffix array two • Repeat Melody 2Time limit:5000mscase time limit:1000msmemory limit:256mb description

Small hi usually a big hobby is playing the piano. We know that a musical melody is represented as a series of numbers consisting of N in length. Little hi has practiced a lot of music and found that many of the works themselves contain the same melody.

Melody can be expressed as a continuous sequence of numbers, similar melodies in the original sequence is not overlapping, for example, 1 2 3 2 3 2 1 in 2 3 2 appeared once, 2 3 appeared two times, little hi want to know a melody in the frequency of at least two times what is the longest melody?

Tips on how to solve problems

Input

The first line is an integer N. 1≤n≤100000

Next there are N integers that represent the number of each tone. 1≤ Digital ≤1000

Output

An integer line that represents the answer.

Sample Input

81 2 3 2 3 2 3 1

Sample Output

2

Analysis:

And POJ1743 is the same ... It's just a matter of two-point judgment. >= and > ...

POJ1743

Code:

#include <algorithm> #include <iostream> #include <cstring> #include <cstdio>//by neighthornusing namespace Std;const int Maxn=100000+5;int N,S[MAXN],GS[MAXN],SA[MAXN],WV[MAXN],WB[MAXN],RAN[MAXN], height[maxn];inline BOOL cmp (int *x,int a,int b,int l) {return x[a]==x[b]&&x[a+l]==x[b+l];} inline void da (int *sa,int *x,int n,int m) {int i,j,p,*y=wb;for (i=0;i<m;i++) gs[i]=0;for (i=0;i<n;i++) gs[x[i]]++; for (i=1;i<m;i++) gs[i]+=gs[i-1];for (i=n-1;~i;i--) sa[--gs[x[i]]]=i;for (j=1,p=1;p<n;j<<=1,m=p) {for (i= n-j,p=0;i<n;i++) y[p++]=i;for (i=0;i<n;i++) if (sa[i]>=j) y[p++]=sa[i]-j;for (i=0;i<n;i++) wv[i]=x[y[i]]; for (i=0;i<m;i++) gs[i]=0;for (i=0;i<n;i++) gs[wv[i]]++;for (i=1;i<m;i++) gs[i]+=gs[i-1];for (i=n-1;~i;i--) Sa[--gs[wv[i]]]=y[i];p =1;swap (x, y), x[sa[0]]=0;for (i=1;i<n;i++) x[sa[i]]=cmp (y,sa[i],sa[i-1],j) p-1:p++;}} inline void calheight (int n) {int i,j,k=0;for (i=0;i<=n;i++) ran[sa[i]]=i;for (i=0;i<n;height[ran[i++]]=k) for (k? K--:233,j=sa[ran[i]-1];s[i+k]==s[j+k];k++);} inline bool Check (int x) {int min=n,max=0;bool ans=0;for (int i=1;i<=n;i++) {if (height[i]<x) {if (max!=0) if (max-min &GT;=X) Ans=1; Min=max=sa[i];continue;} Min=min (Min,sa[i]), Max=max (Max,sa[i]);} if (max!=0&&max-min>=x) Ans=1;return ans;} Signed main (void) {scanf ("%d", &n), for (int i=0;i<n;i++) scanf ("%d", &s[i]), Ran[i]=s[i];d a (sa,ran,n+1,1001 ); Calheight (n); int L=1,r=n>>1,ans=0;while (l<=r) {int mid= (l+r) >>1;if (check (mid)) ans=mid,l=mid+1; Elser=mid-1;} printf ("%d\n", ans); return 0;} The Cap ou pas cap. Pas Cap.

　　

by Neighthorn

Hihocoder #1407: suffix array two • Repeat Melody 2