Repeated Substrings (Bipartite + Suffix Array)

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Title:

Given a string S (S <=1e5) S (| S |<=1e5) S (S <= 1e5), it is required to find the longest substring, and the substring appears more than once and can overlap.

If there are more than one longest substring, the output dictionary order is the smallest.

Train of thought:

For the number of substrings, we first consider the suffix array. We can determine the longest length of the suffix array by the height[i]height[i]height[i] array. Finally, we can use lenlenlenlenlenlenlenlen to scan for the answer with the smallest dictionary order, mainly for the sake of proficiency in the use of suffix arrays.

Code:

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1e5+10;
struct SA{//Panels for suffix arrays
    int sa[N];
    int t1[N],t2[N],c[N];
    int rk[N],ht[N];
    void build(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++)rk[sa[i]]=i;
        for(i=0;i<n;i++)
        {
            if(k)k--;
            j=sa[rk[i]-1];
            while(s[i+k]==s[j+k])k++;
            ht[rk[i]]=k;
        }
    }
}sa;
int s[N];
bool check(int n,int len)//Dichotomy decides whether the current length is legitimate
{
    int cnt=1;
    for(int i=2;i<=n;i++)
    {
        if(sa.ht[i]>=len)
        {
            cnt++;
            if(cnt>=2) return true;//More than once
        }
        else cnt=1;
    }
    return false;
}
char S[N];
int id[N];
int main()
{
    int mx=0;
    scanf("%s",S);
    int n=strlen(S);
    for(int i=0;i<n;i++)
    {
        s[i]=S[i]-'a'+1;
        mx=max(mx,s[i]);
    }
    s[n]=0;//Subscripts start at 0
    sa.build(s,n+1,mx+1);
    sa.getHeight(s,n);
    int l=0,r=n,ans;
    while(l<=r)
    {
        int mid=(l+r)/2;
        if(check(n,mid)) l=mid+1,ans=mid;
        else r=mid-1;
    }
    int pos=0;
    int cnt=1;
    for(int i=2;i<=n;i++)//Find dictionary order minimum
    {
        if(sa.ht[i]>=ans)
        {
            cnt++;pos=i;
            if(cnt>=2){
                break;
            }
        }
        else {
            cnt=1;
        }
    }
    pos=sa.sa[pos];
    for(int i=pos;i<=pos+ans-1;i++)printf("%c",S[i]);puts("");//output
    return 0;
}