poj3261之可重叠的k次最长重复子串

发布时间:2014-10-23 23:22:57
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Milk Patterns Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 8511   Accepted: 3869 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 #include <iostream> #include <cstdio> #include <cstring> #include <string> #include <queue> #include <algorithm> #include <map> #include <iomanip> #define INF 99999999 typedef long long LL; using namespace std; const int MAX=20000+10; int *rank,sa[MAX],height[MAX],temp[MAX]; int wa[MAX],wb[MAX],wm[MAX]; struct Node{ int x,id; bool operator<(const Node &a)const{ if(x == a.x)return id<a.id; return x<a.x; } }r[MAX]; bool cmp(int *r,int a,int b,int l){ return r[a] == r[b] && r[a+l] == r[b+l]; } void makesa(Node *r,int *sa,int n,int m){ int *x=wa,*y=wb,*t,i,j,p; sort(r,r+n);//由于这里最初的m很大所以不采取基数排序,而是采用快速排序,当然也可以先离散化再基数排序 for(i=0;i<n;++i)sa[i]=r[i].id; for(i=1,x[sa[0]]=0;i<n;++i)x[sa[i]]=(r[i].x == r[i-1].x?m-1:m++); /*for(i=0;i<m;++i)wm[i]=0; for(i=0;i<n;++i)wm[x[i]=r[i]]++; for(i=1;i<m;++i)wm[i]+=wm[i-1]; for(i=n-1;i>=0;--i)sa[--wm[x[i]]]=i;*/ for(i=0,j=1,p=0;p<n;j=j*2,m=p){//进行基数排序,当然也可以用快速排序,但是这里基数排序更快 for(p=0,i=n-j;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<m;++i)wm[i]=0; for(i=0;i<n;++i)wm[x[y[i]]]++; for(i=1;i<m;++i)wm[i]+=wm[i-1]; for(i=n-1;i>=0;--i)sa[--wm[x[y[i]]]]=y[i]; for(t=x,x=y,y=t,i=p=1,x[sa[0]]=0;i<n;++i){ x[sa[i]]=cmp(y,sa[i],sa[i-1],j)?p-1:p++; } } rank=x; } void calheight(int *r,int *sa,int n){ for(int i=0,j=0,k=0;i<n;height[rank[i++]]=k){ for(k?--k:0,j=sa[rank[i]-1];r[i+k] == r[j+k];++k); } } int main(){ int n,k; while(cin>>n>>k){ for(int i=0;i<n;++i)cin>>temp[i],r[i].x=temp[i]+1,r[i].id=i; r[n].x=0,r[n].id=n; makesa(r,sa,n+1,1); calheight(temp,sa,n); int l=1,r=n,mid,sum=0,num=1; while(l<=r){ mid=l+r>>1; num=1; for(int i=1;i<=n;++i){ if(height[i]>=mid)++num; else num=1; if(num>=k)break; } if(num>=k)l=mid+1; else r=mid-1; } cout<<r<<endl; } return 0; }

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