POJ - 3261 后缀数组 height应用

题意:求最少重叠\(k\)次的重复子串的最大长度
子串长度问题依然是二分枚举,可以观察出重叠的一定是sa排序中连续的
之前想出一种判断要\(n^2\)的方法,没有考虑到后面肯定会连续出现的情况
(大概想法是枚举重复中的最大\(lcp\)(和之前定义的\(lcp\)有所区别),若存在\(k\)个\((i-j)<=lcp\)既为真←好像很不靠谱的样子)

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<string>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define iin(a) scanf("%d",&a)
#define lin(a) scanf("%lld",&a)
#define din(a) scanf("%lf",&a)
#define s0(a) scanf("%s",a)
#define s1(a) scanf("%s",a+1)
#define print(a) printf("%lld",(ll)a)
#define enter putchar('\n')
#define blank putchar(' ')
#define println(a) printf("%lld\n",(ll)a)
#define IOS ios::sync_with_stdio(0)
using namespace std;
const int maxn = 1e5+11;
const int oo = 0x3f3f3f3f;
const double eps = 1e-7;
typedef long long ll;
ll read(){
    ll x=0,f=1;register char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int str[maxn],n;
struct SA{
    int Rank[maxn],sa[maxn],tsa[maxn],A[maxn],B[maxn];
    int cntA[maxn],cntB[maxn];
    int height[maxn],best[maxn][30],n;//height[i]:第sa[i]与sa[i-1]的cp 
    void get(int nn){
        n=nn; 
        rep(i,0,23333) cntA[i]=0;
        rep(i,1,n) cntA[str[i]]++;
        rep(i,1,23333) cntA[i]+=cntA[i-1];
        rrep(i,n,1) sa[cntA[str[i]]--]=i;
        Rank[sa[1]]=1;
        rep(i,2,n){
            if(str[sa[i]]==str[sa[i-1]]){
                Rank[sa[i]]=Rank[sa[i-1]];
            }else{
                Rank[sa[i]]=1+Rank[sa[i-1]];
            }
        }
        for(int l=1;Rank[sa[n]]<n;l<<=1){
            rep(i,1,n) cntA[i]=cntB[i]=0;
            rep(i,1,n) cntA[A[i]=Rank[i]]++;
            rep(i,1,n) cntB[B[i]=(i+l<=n?Rank[i+l]:0)]++;
            rep(i,1,n) cntA[i]+=cntA[i-1],cntB[i]+=cntB[i-1];
            rrep(i,n,1) tsa[cntB[B[i]]--]=i;
            rrep(i,n,1) sa[cntA[A[tsa[i]]]--]=tsa[i];
            Rank[sa[1]]=1;
            rep(i,2,n){
                bool flag=A[sa[i]]==A[sa[i-1]]&&B[sa[i]]==B[sa[i-1]];
                flag=!flag;
                Rank[sa[i]]=Rank[sa[i-1]]+flag;
            }
        }
    }
    void ht(){
        int j=0;
        rep(i,1,n){
            if(j) j--;
            while(str[i+j]==str[sa[Rank[i]-1]+j]) j++;
            height[Rank[i]]=j;
        }
    }
    void rmq(){
        rep(i,1,n) best[i][0]=height[i];
        for(int i=1;(1<<i)<=n;i++){
            for(int j=1;j+(1<<i)-1<=n;j++){
                best[j][i]=min(best[j][i-1],best[j+(1<<(i-1))][i-1]);
            }
        }
    }
    int query(int l,int r){
        if(l==r)return -oo;
        if(l>r)swap(l,r);
        l++;
        int k=log2(r-l+1);
        return min(best[l][k],best[r-(1<<k)+1][k]);
    }
}sa;
int a[maxn],k;
bool check(int x){
    int cnt=0;bool flag=1;
    for(int i=1;i<=n;i++){
        if(sa.height[i]>=x){
            cnt++;
            if(flag){
            	cnt++;
            	flag=0;
			}
            if(cnt>=k)return 1;
        }else{
			cnt=0;
			flag=1;
        }
    }
    return 0;
}
int main(){
    while(cin>>n>>k){
        rep(i,1,n) str[i]=a[i]=read()+1;
        sort(a+1,a+1+n);
        int m=unique(a+1,a+1+n)-a-1;
        rep(i,1,n) str[i]=lower_bound(a+1,a+1+m,str[i])-a;//保留1 
        str[n+1]=0;
        sa.get(n);
        sa.ht();
        int l=1,r=n,mid,ans=0;
        while(l<=r){
            mid=(l+r)>>1;
            if(check(mid)) l=mid+1,ans=mid;
            else r=mid-1;
        }
        println(ans);
    }
    return 0;
}