poj 3261 Milk Patterns

 

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 19143		Accepted: 8441
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

可重叠的 k 次最长重复子串（pku3261）

给定一个字符串，求至少出现 k 次的最长重复子串，这 k 个子串可以重叠。

算法分析：

这题的做法和上一题差不多，也是先二分答案，然后将后缀分成若干组。不

同的是，这里要判断的是有没有一个组的后缀个数不小于 k。如果有，那么存在

k 个相同的子串满足条件，否则不存在。这个做法的时间复杂度为 O(nlogn)。

 

 

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<cmath>
 5 #include<algorithm>
 6 using namespace std;  
 7 int const N=20000+3; 
 8 int rk[N],r[N],sa[N],num[N*50],wa[N<<1],wb[N<<1],wv[N],n,K,h[N];  
 9 int cmp(int *r,int x,int y,int z){
10     return r[x]==r[y] && r[x+z]==r[y+z];  
11 }
12 void build_sa(int *r,int *sa,int n,int m){
13     int i,j,p,*x=wa,*y=wb;  
14     for(i=0;i<m;i++) num[i]=0;  
15     for(i=0;i<n;i++) num[x[i]=r[i]]++;  
16     for(i=1;i<m;i++) num[i]+=num[i-1];  
17     for(i=n-1;i>=0;i--) sa[--num[x[i]]]=i;  
18     for(p=1,j=1;p<n;j<<=1,m=p){
19         for(p=0,i=n-j;i<n;i++) y[p++]=i;  
20         for(i=0;i<n;i++) if(sa[i]>=j) y[p++]=sa[i]-j;  
21         for(i=0;i<m;i++) num[i]=0;  
22         for(i=0;i<n;i++) wv[i]=x[y[i]]; 
23         for(i=0;i<n;i++) num[wv[i]]++;  
24         for(i=1;i<m;i++) num[i]+=num[i-1];  
25         for(i=n-1;i>=0;i--) sa[--num[wv[i]]]=y[i]; 
26         swap(x,y); 
27         for(p=1,x[sa[0]]=0,i=1;i<n;i++)  
28             x[sa[i]]=cmp(y,sa[i],sa[i-1],j)? p-1:p++;  
29     }
30     for(i=0;i<n;i++) rk[i]=x[i]; 
31 }
32 void build_h(int *r,int *sa,int n){
33     int k=0; 
34     for(int i=0;i<n;i++){
35         if(k) k--; 
36         int j=sa[rk[i]-1];  
37         while (r[i+k]==r[j+k]) k++;  
38         h[rk[i]]=k; 
39     } 
40 }
41 int check(int mid,int n){
42     int num=1;  
43     for(int i=1;i<=n;i++){
44         if(h[i]>=mid){
45             num++;  
46             if(num>=K) return 1; 
47         }else num=1;  
48     }
49     return 0; 
50 }
51     
52 int main(){
53     scanf("%d%d",&n,&K);  
54     for(int i=0;i<n;i++) scanf("%d",&r[i]);  
55     build_sa(r,sa,n+1,1000001); 
56     build_h(r,sa,n);  
57     int l=0,r=n,ans; 
58     while (l<=r){
59         int mid=(l+r)/2;  
60         if(check(mid,n)){
61             ans=mid; 
62             l=mid+1;
63         }else r=mid-1; 
64     }
65     printf("%d\n",ans);  
66     return 0; 
67 }

 

 

 

转载于:https://www.cnblogs.com/ZJXXCN/p/10961710.html