POJ 3261 Milk Patterns 后缀数组求 一个串种 最长可重复子串重复至少k次

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include<vector>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 2e5+10, M = 2e5+20, mod = 1e9+7, inf = 2e9;

///heght[i] 表示 Suffix(sa[i-1])和Suffix(sa[i]) 的最长公共前缀：
///rank[i] 表示 开头为i的后缀的等级：
///sa[i] 表示 排名为i的后缀 的开头位置：

int *rank,r[N],sa[N],height[N],wa[N],wb[N],wm[N];
bool cmp(int *r,int a,int b,int l) {
    return r[a] == r[b] && r[a+l] == r[b+l];
}

void SA(int *r,int *sa,int n,int m) {
    int *x=wa,*y=wb,*t;
    for(int i=0;i<m;++i)wm[i]=0;
    for(int i=0;i<n;++i)wm[x[i]=r[i]]++;
    for(int i=1;i<m;++i)wm[i]+=wm[i-1];
    for(int i=n-1;i>=0;--i)sa[--wm[x[i]]]=i;
    for(int i=0,j=1,p=0;p<n;j=j*2,m=p){
        for(p=0,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)wm[i]=0;
        for(i=0;i<n;++i)wm[x[y[i]]]++;
        for(i=1;i<m;++i)wm[i]+=wm[i-1];
        for(i=n-1;i>=0;--i)sa[--wm[x[y[i]]]]=y[i];
        for(t=x,x=y,y=t,i=p=1,x[sa[0]]=0;i<n;++i) {
            x[sa[i]]=cmp(y,sa[i],sa[i-1],j)?p-1:p++;
        }
    }
    rank=x;
}
void Height(int *r,int *sa,int n) {
    for(int i=0,j=0,k=0;i<n;height[rank[i++]]=k)
    for(k?--k:0,j=sa[rank[i]-1];r[i+k] == r[j+k];++k);
}

int n,a[N],k;
int check(int x) {
        int i = 2,cnt;
        while(1) {
            while(i <= n && height[i] < x) i++;
            if(i > n) break;
            cnt = 1;
            while(i <= n && height[i] >= x) {
                cnt++;
                i++;
            }
            if(cnt >= k) {
                return 1;
            }
        }
        return 0;
}
int main() {
        while(~scanf("%d%d",&n,&k)) {
            for(int i = 1; i <= n; ++i) scanf("%d",&a[i]),r[i-1] = a[i];
            sort(a+1,a+n+1);
            int c = unique(a+1,a+n+1) - a - 1;
            for(int i = 0; i < n; ++i) r[i] = lower_bound(a+1,a+c+1,r[i]) - a;
            r[n] = 0;
            SA(r,sa,n+1,20001);
            Height(r,sa,n);
            int ll = 1, rr = n,ans = 0;
            while(ll <= rr) {
                int md = (ll + rr) >> 1;
                if(check(md)) {
                    ans = md, ll = md + 1;
                } else rr = md - 1;
            }
            printf("%d\n",ans);
        }
        return 0;
}