标签：cas length The pat ons ret lex sch else
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 19664 | Accepted: 8632 | |
| 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
Output
Sample Input
8 2 1 2 3 2 3 2 3 1
Sample Output
4
Source
//#include <bits/stdc++.h> #include <algorithm> #include <cstdio> using namespace std; typedef long long ll; typedef pair<int,int> pii; const int maxn = 1e5 + 10; int n,k,mx; int c[maxn],rk[maxn],sa[maxn],tp[maxn],ton[maxn]; int height[maxn]; inline void rsort(){ for(register int i=0;i<=mx;++i)ton[i]=0; for(register int i=1;i<=n;++i)ton[rk[i]]++; for(register int i=1;i<=mx;++i)ton[i]=ton[i]+ton[i-1]; for(register int i=n;i>=1;--i)sa[ton[rk[tp[i]]]--]=tp[i]; } inline void solve(){ for(register int i=1;i<=n;++i)rk[i]=c[i],tp[i]=i; rsort(); for(register int w=1,p=0;w<=n&&p<n;w*=2,mx=p){ p=0; for(register int i=n-w+1;i<=n;++i){ tp[++p]=i; } for(register int i=1;i<=n;++i){ if(sa[i]>w){ tp[++p]=sa[i]-w; } } rsort(); swap(rk,tp); rk[sa[1]]=p=1; for(register int i=2;i<=n;++i){ rk[sa[i]]=(tp[sa[i-1]]==tp[sa[i]]&&tp[sa[i-1]+w]==tp[sa[i]+w])?p:++p; } if(p>=n)break; } int k=0; for(register int i=1;i<=n;++i)rk[sa[i]]=i; for(register int i=1;i<=n;++i){ if(k)--k; else k=0; int j=sa[rk[i]-1]; while(c[i+k]==c[j+k])++k; height[rk[i]]=k; } } inline bool check(int cur){ int cnt=1; for(register int i=2;i<=n;++i){ if(height[i]<cur)cnt=1; else cnt++; if(cnt>=k)return true; } return false; } int main() { //freopen("1.txt", "r", stdin); scanf("%d%d",&n,&k); for(register int i=1;i<=n;++i){ scanf("%d",&c[i]); ++c[i]; mx=max(mx,c[i]); } //1 2 3 2 3 2 3 1 solve(); int res=0; int l=1,r=n; // for(register int i=1;i<=n;++i){ // printf("debug rk[%d] = %d sa[%d] = %d height[%d] = %d\n",i,rk[i],i,sa[i],i,height[i]); // } while(l<=r){ int mid=l+r>>1; if(check(mid))res=mid,l=mid+1; else r=mid-1; } printf("%d\n",res); return 0; }
标签：cas length The pat ons ret lex sch else
原文地址：https://www.cnblogs.com/czy-power/p/11405834.html
