poj3261 Milk Patterns（后缀数组）

Milk Patterns

Time Limit: 5000MS

 

Memory Limit: 65536K

Total Submissions: 12571

 

Accepted: 5572

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 milkon day i appears on the ith line.

Output

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

Sample Input

8 2

1

2

3

2

3

2

3

1

Sample Output

4

Source

USACO2006 December Gold

 

 

【思路】

       至少出现k次的可重叠最长子串。

       二分长度+划分height，然后判断是否存在一组的数目不小于k即可。

       需要注意的是：用末尾添0的方法解决n==1时的RE问题，但是在can中需要忽略height[0]且考虑height[1..n]。

 

【代码】

 

 1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #define FOR(a,b,c) for(int a=(b);a<=(c);a++) 5 using namespace std; 6  7 const int maxn = 20000+10; 8 const int maxm = 1000000+10; 9 10 int s[maxn];11 int sa[maxn],c[maxm],t[maxn],t2[maxn];12 13 void build_sa(int m,int n) {14     int i,*x=t,*y=t2;15     for(i=0;i<m;i++) c[i]=0;16     for(i=0;i<n;i++) c[x[i]=s[i]]++;17     for(i=1;i<m;i++) c[i]+=c[i-1];18     for(i=n-1;i>=0;i--) sa[--c[x[i]]]=i;19     20     for(int k=1;k<=n;k<<=1) {21         int p=0;22         for(i=n-k;i<n;i++) y[p++]=i;23         for(i=0;i<n;i++) if(sa[i]>=k) y[p++]=sa[i]-k;24         25         for(i=0;i<m;i++) c[i]=0;26         for(i=0;i<n;i++) c[x[y[i]]]++;27         for(i=0;i<m;i++) c[i]+=c[i-1];28         for(i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];29         30         swap(x,y);31         p=1; x[sa[0]]=0;32         for(i=1;i<n;i++) 33             x[sa[i]]=y[sa[i]]==y[sa[i-1]] && y[sa[i]+k]==y[sa[i-1]+k]?p-1:p++;34         if(p>=n) break;35         m=p;36     }37 }38 int rank[maxn],height[maxn];39 void getHeight(int n) {40     int i,j,k=0;41     for(i=0;i<=n;i++) rank[sa[i]]=i;42     for(i=0;i<n;i++) {43         if(k) k--;44         j=sa[rank[i]-1];45         while(s[j+k]==s[i+k]) k++;46         height[rank[i]]=k;47     }48 }49 50 int n,k;51 52 bool can(int limit) {53     int cnt=1;                //忽略height[0] 截至height[n]54     for(int i=2;i<=n;i++) {    55         if(height[i]>=limit) {56             if(++cnt>=k) return true;57         }58         else cnt=1;59     }60     return false;61 }62 63 int main() {64     scanf("%d%d",&n,&k);65     int up=0;66     FOR(i,0,n-1) {67          scanf("%d",&s[i]);68          s[i]++;69          up=max(up,s[i]);70     }71     s[n]=0;72     build_sa(up+1,n);73     getHeight(n);74     int L=0,R=n;75     while(L<R) {76         int M=L+(R-L+1)/2;77         if(can(M)) L=M;78         else R=M-1;79     }80     printf("%d\n",L);81     return 0;82 }