[BZOJ1717][Usaco2006 Dec]Milk Patterns 产奶的模式（后缀数组）

题目描述

题解

代码

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

const int max_n=2e4+5;
const int max_m=1e6+5;

int n,k,l,r,mid,ans,m=1e6+1;
int s[max_n];
int sa[max_n],height[max_n],rank[max_n],c[max_m],x[max_n],y[max_n];

inline void build_sa()
{
for (int i=0;i<m;++i) c[i]=0;
for (int i=0;i<n;++i) c[x[i]=s[i]]++;
for (int i=1;i<m;++i) c[i]+=c[i-1];
for (int i=n-1;i>0;--i) sa[--c[x[i]]]=i;
for (int k=1;k<=n;k<<=1)
{
int p=0;
for (int i=n-k;i<n;++i) y[p++]=i;
for (int i=0;i<n;++i) if (sa[i]>=k) y[p++]=sa[i]-k;

for (int i=0;i<m;++i) c[i]=0;
for (int i=0;i<n;++i) c[x[y[i]]]++;
for (int i=1;i<m;++i) c[i]+=c[i-1];
for (int i=n-1;i>=0;--i) sa[--c[x[y[i]]]]=y[i];

swap(x,y);
p=1; x[sa[0]]=0;
for (int i=1;i<n;++i)
x[sa[i]]=(y[sa[i-1]]==y[sa[i]]&&
(sa[i-1]+k>=n?-1:y[sa[i-1]+k])==(sa[i]+k>=n?-1:y[sa[i]+k]))?p-1:p++;
if (p>=n) break;
m=p;
}
}
inline void build_lcp()
{
for (int i=0;i<n;++i) rank[sa[i]]=i;
int k=0;
height[0]=0;
for (int i=0;i<n;++i)
{
if (!rank[i]) continue;
if (k) --k;
int j=sa[rank[i]-1];
while (i+k<n&&j+k<n&&s[i+k]==s[j+k]) ++k;
height[rank[i]]=k;
}
}
inline bool check(int x)
{
int num=0;
for (int i=0;i<n;++i)
if (height[i]>=x)
{
num++;
if (num==k-1) return true;
}
else num=0;
return false;
}
int main()
{
scanf("%d%d",&n,&k);
for (int i=0;i<n;++i) scanf("%d",&s[i]);
build_sa();
build_lcp();
l=1; r=n;
while (l<=r)
{
mid=(l+r)>>1;
if (check(mid)) ans=mid,l=mid+1;
else r=mid-1;
}
printf("%d\n",ans);
}