poj3261(后缀数组+二分)
来源:互联网 发布:普林科技 大数据 编辑:程序博客网 时间:2024/06/06 01:29
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 overN (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 leastK times.
Input
Lines 2..N+1: N integers, one per line, the quality of the milk on dayi appears on theith line.
Output
Sample Input
8 212323231
Sample Output
4
Source
#include<iostream>#include<cstdio>#include<cstring>using namespace std;//*****************************************************************const int MAXN=1000000+100;int str[MAXN];//待处理字符串int sa[MAXN];//求得的后缀数组int wa[MAXN],wb[MAXN],wv[MAXN],wh[MAXN];int cmp(int *r,int a,int b,int l){return r[a]==r[b]&&r[a+l]==r[b+l];}//求后缀数组sa[],下标1到n-1(此处n=strlen(str)+1)有效后缀//将str的n个后缀从小到大进行排序之后把排好序的后缀的开头位置顺次放入sa中。//保证Suffix(sa[i])<Suffix(sa[i+1])//1<=i<n,sa[0]存放人为添加在末尾的那个最小的后缀//倍增算法的时间复杂度为O(nlogn)//倍增算法的空间复杂度都是O(n)void da(int *r,int *sa,int n,int m){int i,j,p,*x=wa,*y=wb,*t;for(i=0;i<m;i++) wh[i]=0;for(i=0;i<n;i++) wh[x[i]=r[i]]++;for(i=1;i<m;i++) wh[i]+=wh[i-1];for(i=n-1;i>=0;i--) sa[--wh[x[i]]]=i;for(j=1,p=1;p<n;j*=2,m=p){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<n;i++) wv[i]=x[y[i]];for(i=0;i<m;i++) wh[i]=0;for(i=0;i<n;i++) wh[wv[i]]++;for(i=1;i<m;i++) wh[i]+=wh[i-1];for(i=n-1;i>=0;i--) sa[--wh[wv[i]]]=y[i];for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1;i<n;i++)x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;}return;}int rank[MAXN],height[MAXN];//定义height[i]=suffix(sa[i-1])和suffix(sa[i])的最长公//共前缀,也就是排名相邻的两个后缀的最长公共前缀//任意两个起始位置为i,j(假设rank[i]<rank[j])的后缀的最长公共前缀//为height[rank[i]+1]、height[rank[i]+2]…height[rank[j]]的最小值void calheight(int *r,int *sa,int n){int i,j,k=0;for(i=1;i<=n;i++) rank[sa[i]]=i;for(i=0;i<n;height[rank[i++]]=k)for(k?k--:0,j=sa[rank[i]-1];r[i+k]==r[j+k];k++);return;}int Max(int a,int b){return a<b?b:a;}int Solve(int num,int tim){int i,Maxlen,Minlen,midlen;Minlen=0,Maxlen=num;while(Minlen<=Maxlen){midlen=(Minlen+Maxlen)/2;int tmp=1;for(i=1;i<=num;i++){if(height[i]<midlen){tmp=1;}else{tmp++;}if(tmp>=tim)break;}if(tmp>=tim)Minlen=midlen+1;else Maxlen=midlen-1;}return Maxlen;}int main(){int num,tim,i;//freopen("in.txt","r",stdin);while(~scanf("%d%d",&num,&tim)){for(i=0;i<num;i++){scanf("%d",&str[i]);str[i]++;//保证在末尾差的0是最小的}str[num]=0;da(str,sa,num+1,MAXN);calheight(str,sa,num);printf("%d\n",Solve(num,tim));}return 0;}
- poj3261(后缀数组+二分)
- 后缀数组+二分poj3261
- poj3261(后缀数组+二分)
- POJ3261---Milk Patterns(后缀数组+二分)
- 【POJ3261】Milk Patterns【后缀数组】【二分】
- 【poj3261】Milk Patterns 后缀数组+二分
- poj3261 Milk Patterns 后缀数组+二分
- POJ3261 Milk Patterns(后缀数组,二分)
- POJ3261----后缀数组
- POJ3261之后缀数组
- 【后缀数组】Milk Patterns POJ3261
- poj3261 Milk Patterns 后缀数组
- 【POJ3261】Milk Patterns 后缀数组
- POJ3261:Milk Patterns(后缀数组)
- 后缀数组 - poj3261 Milk Patterns
- 【POJ3261】 Milk Patterns 后缀数组
- [后缀数组]poj3261 Milk Patterns
- POJ3261 Milk Patterns 【后缀数组】
- JS 判断字符串中是否包含中文
- PowerDesigner 15 导入表
- Liferay工程调用RoleLocalServiceUtil方法报no hibernate session bound to Thread 异常的解决方法
- oracle数据的恢复与备份
- 如何在程序启动默认浏览器与电子邮件系统
- poj3261(后缀数组+二分)
- webdriver知识点总结
- JSON 教程
- Android 学习笔记 Contacts (二)Contacts 联系人详解
- vc中用WaveOut写音乐播放器
- android listView嵌套gridview的使用心得
- redhat centos linux防火墙配置
- Hibernate的一级缓存简介
- mtd-utils 及 ubi-utils 交叉编译 .