poj 3261 Milk Patterns

Milk Patterns

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 4698		Accepted: 2012
Case Time Limit: 2000MS

Description

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

To perform a rigorous study, he has invented a complex classification scheme by which each milk sample is recorded as an integer between 0 and 1,000,000 inclusive, and has 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 may 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 that at least one subsequence is repeated at least K times.

Input

Line 1: Two space-separated integers:  N and  K 
Lines 2.. N+1:  N integers, one per line, the quality of the milk on day  i appears on the  ith line.

Output

Line 1: One integer, the length of the longest pattern which occurs at least  K times

Sample Input

8 2
1
2
3
2
3
2
3
1

Sample Output

4

Source

USACO 2006 December Gold

 

/* 可重叠的k次最长重复子串 二分即可 */ #include #include #include using namespace std; const int maxn=1010000; char str[maxn]; int wa[maxn],wb[maxn],wv[maxn],wn[maxn],a[maxn],sa[maxn]; int cmp(int* r,int a,int b,int l) { return r[a]==r[b]&&r[a+l]==r[b+l]; } //n为字符串长度，m为字符的取值范围，r为字符串。后面的j为每次排序时子串的长度 void DA(int* r,int* sa,int n,int m) { int i,j,p,*x=wa,*y=wb,*t; ///对R中长度为1的子串进行基数排序 for(i=0; i=0; i--)sa[--wn[x[i]]]=i; for(j=1,p=1; p=j)y[p++]=sa[i]-j; ///基数排序 for(i=0; i=0; i--)sa[--wn[wv[i]]]=y[i]; ///当p=n的时候，说明所有串都已经排好序了 ///在第一次排序以后，rank数组中的最大值小于p，所以让m=p for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1; in //for(int i=1; i>1; // printf("mid=%d/n",mid); bool flag=false; int num=0; for(int i=1; i<=n; i++) if(height[i]>=mid) num++; else { if(num+1>=k) { ans=mid; flag=true; } num=0; } if(num+1>=k) { ans=mid; flag=true; } if(flag) fir=mid+1; else sec=mid-1; } printf("%d/n",ans); } return 0; }