POJ 3261

Milk Patterns

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 11073		Accepted: 4949
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

Sample Output

4
找一个子串，至少出现了k次，而且长度最长
先二分枚举子串的长度，然后再hash,看是否出现了k次

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<string.h>
 4 #include<algorithm>
 5 #include<map>
 6 using namespace std;
 7 typedef unsigned long long ull;
 8 #define mmax 20000+10
 9 ull ansx[mmax],ansy[mmax],base[mmax],seed=131;
10 int a[mmax];
11 int n,k;
12 bool judge(int len){
13     memset(ansx,0,sizeof(ansx));
14     for(int i=0;i<len;i++) ansx[0]=ansx[0]*seed+a[i] ;
15     for(int i=len;i<n;i++) {
16         ansx[i-len+1]=ansx[i-len]*seed-(a[i-len] )*base[len]+a[i] ;
17     }
18     sort(ansx,ansx+n-len+1);
19     int cnt=1;
20     for(int i=len;i<n;i++) {
21         if(ansx[i-len+1]==ansx[i-len]) cnt++;
22         else cnt=1;
23         if(cnt>=k) return 1;
24     }
25     return 0;
26 }
27 int solve(){
28     int left=0,right=n,mid,ans=0;
29     while(left<=right){
30         mid=(left+right)>>1;
31         if(judge(mid)) ans=mid,left=mid+1;
32         else right=mid-1;
33     }
34     return ans;
35 }
36 int main(){
37     base[0]=1;
38     for(int i=1;i<=20000;i++) base[i]=base[i-1]*seed;
39     while(scanf("%d%d",&n,&k)!=EOF){
40         for(int i=0;i<n;i++) scanf("%d",&a[i]);
41         printf("%d
",solve());
42     }
43 }