[BZOJ3998] [TJOI2015] string theory (suffix array)

Face questions

What is for a given string of length N, find its first string K kid.

T 0 indicates the same positions in different sub-strings of a count. T = 1 indicates a different position of the same sub-strings plurality counted. K significance as the title.

analysis

Different positions of the same sub-strings counted as one (T = 0)

Each sub-string is a prefix of suffix. Then we enumerate the suffix i ranked according to rank order. Then the next extension will produce \ (n-sa [i] +1 \) substrings. And sub-strings lexicographically than \ (sa [i-1] \) is large, or they will be smaller as a prefix top surface. Considering the \ (height [I] = \ {text} LCP (SA [-I. 1], SA [I]) \) , that is to say with a \ (height [i] \) sub-string is repeated.

Thus, each has a suffix \ (n-sa [i] + 1-height [i] \) of different nature substrings. We have been accumulating this number, accumulated into \ (K \) stop when you can.

A plurality of the same substring count (T = 1) at different locations

Record \ (sum [i] \) represents \ (sa [1], sa [2] \ dots sa [i] \) of the substring and the number of, different positions of the same sub-strings plurality counted. Easy to find \ (sum [i] = \ sum_ {j = 1} ^ i (n-sa [j] +1) \)

We enumerate the first substring \ (i \) character on the position \ (J \) . Then the first substring where the suffix \ (I \) position has been determined, the following character can also be selected, according to the lexicographical sort must be in a continuum. The \ (I =. 3, J = 'B' \) :

aabc
aabcdaabc

Two separated at the beginning of i-th bit substring where the suffix j rank interval \ ([L_2, R_2] \) , then the number of such sub-string \ (cnt = sum [r_2] -sum [l_2-1] - ( L_2 +. 1-R_2) (. 1-i) \) . wherein \ ((r_2-l_2 + 1 ) (i-1) \) subtracting i is less than the length of the substring.

And to the second \ (I \) bits to the end of the substring with a \ (r_2-l_2 + 1 \ ) a. When \ (cnt> K \) when it is determined that this letter, otherwise k minus, continue to enumerate the next letter.

For details, see code.

Code


#include<iostream>
#include<cstdio>
#include<cstring>
#define maxn 500000
#define maxs 128 
using namespace std;
typedef long long ll;

void rsort(int *ans,int *fi,int *se,int sz,int maxv){
    static int buck[maxn+5];
    for(int i=0;i<=maxv;i++) buck[i]=0;
    for(int i=1;i<=sz;i++) buck[fi[i]]++;
    for(int i=1;i<=maxv;i++) buck[i]+=buck[i-1];
    for(int i=sz;i>=1;i--) ans[buck[fi[se[i]]]--]=se[i];
}
int sa[maxn+5],rk[maxn+5],height[maxn+5];
void suffix_sort(char *str,int n){
    static int se[maxn+5];
    for(int i=1;i<=n;i++){
        rk[i]=str[i];
        se[i]=i;
    }
    rsort(sa,rk,se,n,maxs);
    for(int k=1,m=maxs;k<=n;k*=2){
        int p=0;
        for(int i=n-k+1;i<=n;i++) se[++p]=i;
        for(int i=1;i<=n;i++){
            if(sa[i]>k) se[++p]=sa[i]-k;
        }
        rsort(sa,rk,se,n,m);
        swap(rk,se);
        p=rk[sa[1]]=1;
        for(int i=2;i<=n;i++){
            if(se[sa[i-1]]==se[sa[i]]&&se[sa[i-1]+k]==se[sa[i]+k]) rk[sa[i]]=p;
            else rk[sa[i]]=++p;
        }
        if(p==n) break;
        m=p;
    }
}
void get_height(char *str,int n){
    suffix_sort(str,n);
    for(int i=1;i<=n;i++) rk[sa[i]]=i;
    int k=0;
    for(int i=1;i<=n;i++){
        if(k) k--;
        int j=sa[rk[i]-1];
        while(str[i+k]==str[j+k]) k++;
        height[rk[i]]=k;
    }
}

int n,T,K;
char s[maxn+5];
void solve0(){
    int now=0,last=0;
    for(int i=1;i<=n;i++){
        now=last+n-sa[i]+1-height[i];
        if(now>K){
            for(int j=0;j<K+height[i]-last;j++){
                putchar(s[sa[i]+j]);
            }
            return;
        }
        last=now;
    }
    printf("-1\n");
}

int bin_search(int l,int r,int len,char ch){
    while(l<=r){
        int mid=(l+r)>>1;
        if(s[sa[mid]+len-1]>ch){
            r=mid-1;
        }else l=mid+1;
    }
    return r;
}
 
void solve1(){
    //[Warning] You are not expected to understand this.
    static ll sum[maxn+5];//后缀sa[1],sa[2],...sa[i]的所有不同子串个数之和 
    for(int i=1;i<=n;i++) sum[i]=sum[i-1]+n-sa[i]+1; 
    if(K>sum[n]){
        printf("-1\n");
        return; 
    }
    int l1=1,r1=n;
    for(int i=1;i<=n;i++){//枚举子串的第i位的字符
        int l2=l1;
        for(char j='a';j<='z';j++){//按字典序枚举 
            int r2=bin_search(l2,r1,i,j);//二分出第i位为j的子串开头所在后缀的rank区间[l2,r2] 
            ll cnt=sum[r2]-sum[l2-1]-(ll)(r2-l2+1)*(i-1); //第i位为j的子串个数,减去(i-1)*(r2-l2+1)是把长度不足i的子串减掉 
            if(K<=cnt){//剩下的个数比cnt小，说明子串的第i位一定是j 
                if(K<=r2-l2+1){
                    //第i位是j的子串中,最小的r2-l2+1个子串显然结尾是j 
                    //如果满足这个条件,那子串的结尾一定是j,后面没有其他字符,可以直接输出
                     for(int u=sa[l2];u<=sa[l2]+i-1;u++) putchar(s[u]);
                     return; 
                }else{
                    l1=l2;
                    r1=r2;
                    K-=r2-l2+1;
                    break;
                    //否则这个子串第i位是j,后面还有其他字符,所以要break掉再去枚举i+1位 
                }
            }else{
                //第i位应该大于j 
                l2=r2+1;//枚举下一个rank区间
                K-=cnt; //减去当前已经算过的个数 
            } 
        } 
        if(n-sa[l1]+1==i) l1++; //后缀sa[l1]的长度只有i,无法枚举到i+1位，所以去掉l1 
    } 
}
int main(){
    scanf("%s",s+1);
    scanf("%d %d",&T,&K);
    n=strlen(s+1);
    get_height(s,n);    
    if(T==0){
        solve0(); 
    }else{
        solve1();
    }
}