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| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 12128 | Accepted: 5387 | |
| 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
ac代码
Problem: 3261 User: kxh1995 Memory: 680K Time: 47MS Language: C++ Result: Accepted
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
int s[20020];
int sa[20020],t1[20020],t2[20020],c[20020];
int rank[20020],height[20020];
void build_sa(int s[],int n,int m)
{
int i,j,p,*x=t1,*y=t2;
for(i=0;i<m;i++)
c[i]=0;
for(i=0;i<n;i++)
c[x[i]=s[i]]++;
for(i=1;i<m;i++)
c[i]+=c[i-1];
for(i=n-1;i>=0;i--)
sa[--c[x[i]]]=i;
for(j=1;j<=n;j<<=1)
{
p=0;
for(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++)
c[i]=0;
for(i=0;i<n;i++)
c[x[y[i]]]++;
for(i=1;i<m;i++)
c[i]+=c[i-1];
for(i=n-1;i>=0;i--)
sa[--c[x[y[i]]]]=y[i];
swap(x,y);
p=1;
x[sa[0]]=0;
for(i=1;i<n;i++)
x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+j]==y[sa[i]+j]?p-1:p++;
if(p>=n)
break;
m=p;
}
}
void getHeight(int s[],int n)
{
int i,j,k=0;
for(i=0;i<=n;i++)
rank[sa[i]]=i;
for(i=0;i<n;i++)
{
if(k)
k--;
j=sa[rank[i]-1];
while(s[i+k]==s[j+k])
k++;
height[rank[i]]=k;
}
}
int judge(int n,int k,int len)
{
int num=1;
for(int i=1;i<=n;i++)
{
if(height[i]>=len)
{
num++;
if(num>=k)
return 1;
}
else
num=1;
}
return 0;
}
int main()
{
int n,k;
while(scanf("%d%d",&n,&k)!=EOF)
{
int i;
int num=-1;
for(i=0;i<n;i++)
{
scanf("%d",&s[i]);
if(num<s[i])
num=s[i];
}
s[n]=0;
build_sa(s,n+1,num+1);
getHeight(s,n);
int ans=0;
int l=0,r=n;
while(l<=r)
{
int mid=(l+r)>>1;
if(judge(n,k,mid))
{
ans=mid;
l=mid+1;
}
else
r=mid-1;
}
printf("%d\n",ans);
}
}
版权声明：本文为博主原创文章，未经博主允许不得转载。
POJ 题目3261 Milk Patterns（后缀数组求最长重叠至少k次的子串长度）
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原文地址：http://blog.csdn.net/yu_ch_sh/article/details/47756337
