hdu 6704 K-th occurrence(后缀数组+可持久化线段树)

Problem Description

You are given a string S consisting of only lowercase english letters and some queries.

For each query (l,r,k), please output the starting position of the k-th occurence of the substring SlSl+1...Sr in S.

Input

The first line contains an integer T(1≤T≤20), denoting the number of test cases.

The first line of each test case contains two integer N(1≤N≤105),Q(1≤Q≤105), denoting the length of S and the number of queries.

The second line of each test case contains a string S(|S|=N) consisting of only lowercase english letters.

Then Q lines follow, each line contains three integer l,r(1≤l≤r≤N) and k(1≤k≤N), denoting a query.

There are at most 5 testcases which N is greater than 103.

Output

For each query, output the starting position of the k-th occurence of the given substring.

If such position don't exists, output −1 instead.

Sample Input

2

12 6

aaabaabaaaab

3 3 4

2 3 2

7 8 3

3 4 2

1 4 2

8 12 1

1 1

a

1 1 1

Sample Output

5

2

-1

6

9

8

1

思路：

我们要找l~r的字串的兄弟串不难想到求lcp 通过二分我们可以找到最右和最左的排名 然后我们只要用可持久线段树维护下标  然后求下标第k小的字符串即可

#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const int N = 1e5+1000;
const int inf = 0x3f3f3f3f;
const double eps = 1e-6;
typedef long long ll;
const ll mod = 1e9+7;
int rt[N];
struct tree{
    int l,r,v,ls,rs;
}t[N<<5];
int nico;
void build(int &p,int l,int r){
    p=++nico;
    t[p].l=l; t[p].r=r;
    if(l==r){
        return ;
    }
    int mid=(l+r)>>1;
    build(t[p].ls,l,mid);
    build(t[p].rs,mid+1,r);
}
void update(int &p,int last,int x){
    p=++nico;
    t[p]=t[last];
    t[p].v++;
    if(t[p].l==t[p].r&&t[p].l==x){
        t[p].v=1;
        return ;
    }
    int mid=(t[p].l+t[p].r)>>1;
    if(x<=mid) update(t[p].ls,t[last].ls,x);
    else update(t[p].rs,t[last].rs,x);
}
int query(int p,int last,int k){
    if(t[p].l==t[p].r){
        return t[p].l;
    }
    int tmp=t[t[p].ls].v-t[t[last].ls].v;
    int mid=(t[p].l+t[p].r)>>1;
    if(tmp>=k) return query(t[p].ls,t[last].ls,k);
    else return query(t[p].rs,t[last].rs,k-tmp);
}
struct S_array{
    int s[N],sa[N],t[N],t2[N],c[N],n;
    int f[N][20];
    void build_sa(int m){
        int i,*x=t,*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(int k=1;k<=n;k<<=1){
            int p=0;
            for(i=n-k;i<n;i++)y[p++]=i;
            for(i=0;i<n;i++)if(sa[i]>=k)y[p++]=sa[i]-k;
            for(i=0;i<m;i++)c[i]=0;
            for(i=0;i<n;i++)c[x[y[i]]]++;
            for(i=0;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]+k]==y[sa[i]+k]?p-1:p++;
            if(p>=n)break;
            m=p;
        }
    }
    int rank[N],height[N];
    void getHeight(){
        int i,j,k=0;
        for(i=0;i<n;i++)rank[sa[i]]=i;
        for(i=0;i<n;i++){
            if(k)k--;
            int j=sa[rank[i]-1];
            while(s[i+k]==s[j+k])k++;
            height[rank[i]]=k;
        }
    }
    void rmq(){
        for(int i=1;i<n;i++)    f[i][0]=height[i];
        for(int j=1;j<20;j++)
            for(int i=1;i+(1<<j)-1<n;i++)
                f[i][j]=min(f[i][j-1],f[i+(1<<(j-1))][j-1]);
    }
    int lcp(int l,int r){
        int k=log2(r-l+1);
        return min(f[l][k],f[r+1-(1<<k)][k]);
    }
    void bu(){
        for(int i=1;i<n;i++)
            update(rt[i],rt[i-1],sa[i]);
    }
    int work(int l,int r,int k){
        int po=rank[l-1];
        //cout<<po<<endl;
        int L,R,ans=-1; int mxl,mxr;
        L=1; R=po; mxl=mxr=po;
        while(L<=R){
            int mid=(L+R)>>1;
        //    cout<<L<<" "<<R<<" "<<mid<<" "<<lcp(mid,po)<<endl;
            if(lcp(mid,po)>=r-l+1){
                R=mid-1;
                ans=mid;
            }else{
                L=mid+1;
            }
        }
        if(ans!=-1)
        mxl=ans;
        ans=-1;
        if(po<n-1){
            L=po+1; R=n-1;
            while(L<=R){
            //    cout<<L<<" "<<R<<endl;
                int mid=(L+R)>>1;
                if(lcp(po+1,mid)>=r-l+1){
                    L=mid+1;
                    ans=mid;
                }else{
                    R=mid-1;
                }
            }
            if(ans!=-1)
            mxr=ans;
        }
    //    cout<<n<<endl;
    //    cout<<mxl<<" "<<mxr<<endl;
        if(lcp(mxl,mxr)>=r-l+1){
            mxl=mxl-1;    
        }
        if(mxr-mxl+1<k) return -1;
        return query(rt[mxr],rt[mxl-1],k)+1;
    }
}sa;
char s[N];
int main(){
//    ios::sync_with_stdio(false);
//    cin.tie(0); cout.tie(0);
    int t; scanf("%d",&t);
    while(t--){
        int n,q; scanf("%d%d",&n,&q);
        scanf("%s",s);
        nico=0;
        for(int i=0;i<n;i++)
            sa.s[i]=s[i]-'a'+1;
        sa.s[n]=0; sa.n=n+1;
        sa.build_sa(27); sa.getHeight();
        build(rt[0],0,N);
        sa.rmq(); sa.bu();
        for(int i=1;i<=q;i++){
            int l,r,k;
            scanf("%d%d%d",&l,&r,&k);
            printf("%d
",sa.work(l,r,k));
        }
    }
}