zoukankan      html  css  js  c++  java
  • Milk Patterns poj3261(后缀数组)

    Milk Patterns
    Time Limit: 5000MS   Memory Limit: 65536K
    Total Submissions: 9274   Accepted: 4173
    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

     1 #include <iostream>
     2 #include <stdio.h>
     3 #include <string.h>
     4 using namespace std;
     5 #define N 20001
     6 #define M 1000002
     7 int a[N],c[N],d[N],e[N],sa[N],height[N],n,b[M],m,k;
     8 int cmp(int *r,int a,int b,int l)
     9 {
    10     return r[a]==r[b]&&r[a+l]==r[b+l];
    11 }
    12 void da()
    13 {
    14     int i,j,p,*x=c,*y=d,*t;
    15     memset(b,0,sizeof(b));
    16     for(i=0; i<n; i++)b[x[i]=a[i]]++;
    17     for(i=1; i<m; i++)b[i]+=b[i-1];
    18     for(i=n-1; i>=0; i--)sa[--b[x[i]]]=i;
    19     for(j=1,p=1; p<n; j*=2,m=p)
    20     {
    21         for(p=0,i=n-j; i<n; i++)y[p++]=i;
    22         for(i=0; i<n; i++)if(sa[i]>=j)y[p++]=sa[i]-j;
    23         for(i=0; i<n; i++)e[i]=x[y[i]];
    24         for(i=0; i<m; i++)b[i]=0;
    25         for(i=0; i<n; i++)b[e[i]]++;
    26         for(i=1; i<m; i++)b[i]+=b[i-1];
    27         for(i=n-1; i>=0; i--)sa[--b[e[i]]]=y[i];
    28         for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1; i<n; i++)
    29             x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;
    30     }
    31 }
    32 void callheight()
    33 {
    34     int i,j,k=0;
    35     b[0]=0;
    36     for(i=1; i<n; i++)b[sa[i]]=i;
    37     for(i=0; i<n-1; height[b[i++]]=k)
    38         for(k?k--:0,j=sa[b[i]-1]; a[i+k]==a[j+k]; k++);
    39 }
    40 bool check(int mid)
    41 {
    42     int sum,i=1;
    43     while(1)
    44     {
    45         while(i<n&&height[i]<mid)i++;
    46         if(i>=n)return 0;
    47         sum=1;
    48         while(i<n&&height[i]>=mid)
    49         {
    50             sum++;
    51             i++;
    52         }
    53         if(sum>=k)return 1;
    54     }
    55 }
    56 int main()
    57 {
    58     int i,l,r;
    59     scanf("%d%d",&n,&k);
    60     for(i=0; i<n; i++)scanf("%d",&a[i]),a[i]++,m=m<a[i]?a[i]:m;
    61     m=m<n?n:m;
    62     m++;
    63     a[n++]=0;
    64     da();
    65     callheight();
    66     l=0,r=n;
    67     while(l<=r)
    68     {
    69         int mid=(l+r)>>1;
    70         if(check(mid))
    71             l=mid+1;
    72         else r=mid-1;
    73     }
    74     printf("%d
    ",r);
    75 }
    View Code
  • 相关阅读:
    C# MVC解决跨站请求伪造(appscan)
    .net中关于Url传参问题
    二月项目完成小结
    sql 获取时间
    ajax提交form表单
    C# 视图遍历List数组
    C#遍历指定文件夹中的所有文件
    C#关于文件的操作
    .net 文件上传到服务器【转】
    Server.MapPath获取各级目录【转】
  • 原文地址:https://www.cnblogs.com/ERKE/p/3593785.html
Copyright © 2011-2022 走看看