bzoj1717: [Usaco2006 Dec]Milk Patterns 产奶的模式 后缀数组

后缀数组求至少重复k次的最长字符串，二分分组即可

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n,m,p[300010],height[300010],sa[300010],str[300010],cnt[1000010],rank[300010],temp[300010];
void init()
{
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++)
    scanf("%d",&str[i]);
}
inline bool epu(int x,int y,int l){return rank[x]==rank[y]&&rank[x+l]==rank[y+l];}
void doubling()
{
    for(int i=0;i<=n;i++)sa[i]=i,rank[i]=str[i];
    for(int i,l=0,pos=0,sig=1000010;pos<n-1;sig=pos)
    {
        for(i=n-l,pos=0;i<n;i++)p[pos++]=i;
        for(i=0;i<n;i++)if(sa[i]>=l)p[pos++]=sa[i]-l;
        memset(cnt,0,sizeof(cnt));
        for(i=0;i<n;i++)cnt[rank[i]]++;
        for(i=1;i<=sig;i++)cnt[i]+=cnt[i-1];
        for(i=n-1;i>=0;i--)sa[--cnt[rank[p[i]]]]=p[i];
        for(temp[sa[0]]=pos=0,i=1;i<n;i++)
        {
            if(!epu(sa[i],sa[i-1],l))pos++;
            temp[sa[i]]=pos;
        }
        for(i=0;i<n;i++)rank[i]=temp[i];
        if(l==0)str[n]=rank[n]=-1;
        if(!l)l=1;else l<<=1;
    }
     
}
void pre()
{
    int i,k;
    for(i=k=0;i<n;i++)
    {
        if(k)k--;
        if(rank[i]==0)continue;
        for(int j=sa[rank[i]-1];str[i+k]==str[j+k];)
            k++;
        height[rank[i]]=k;
    }
}
bool pd(int pp)
{
    int ans=1;
    for(int i=1;i<n;i++)
    {
        if(height[i]<pp)ans=1;
        else ans++;
        if(ans>=m)return true;
    }
    return false;
}
void work()
{
    int l=0,r=n,mid;
    while(l+1<r)
    {
        mid=(l+r)>>1;
        if(pd(mid))l=mid;
        else r=mid;
    }
    if(pd(r))printf("%d\n",r);
    else printf("%d\n",l);
}
int main()
{
    //freopen("xf.in","r",stdin);
    scanf("%d%d",&n,&m);
    memset(str,0,sizeof(str));
    memset(height,0,sizeof(height));
    memset(cnt,0,sizeof(cnt));
    memset(rank,0,sizeof(rank));
    memset(temp,0,sizeof(temp));
    memset(p,0,sizeof(p));
    memset(sa,0,sizeof(sa));
    for(int i=0;i<n;i++)
    scanf("%d",&str[i]);
    doubling();
    pre();
    work();
    return 0;
}