2020牛客暑期多校训练营（第一场）A.B-Suffix Array

2020牛客暑期多校训练营（第一场）A.B-Suffix Array

题目链接

题目描述

The B-function $B(t_1{t}_2\dots t_k)=b_1{b}_2\dots b_k$ of a string $t_1{t}_2\dots t_k$ is defined as follows.

• If there is an index j < i where $t_j=t_i$, $b_i={\min }_{1\le j<i,t_j=t_i}\{i-j\}$
• Otherwise, $b_i=0$.

Given a string $s_1{s}_2\dots s_n$, sort its $n$ suffixes into increasing lexicographically order of the B-function.

Formally, the task is to find a permutaion $p_1,p_2,\dots ,p_n$ of $\{1,2,\dots ,n\}$ such that $B(s_{p_{i-1}}\dots s_n)<B(s_{p_i}\dots s_n)$ holds for $i=2,\dots ,n$.

输入描述:

The input consists of several test cases terminated by end-of-file.
The first line of each test case contains an integer n.
The second line contains a string $s_1{s}_2\dots s_n$.

• $1\le n\le 10^5$
• $s_i$ is either ‘a’ or ‘b’
• The sum of n does not exceed $10^6$

输出描述:

For each test case, print n integers which denote the result.

示例1

输入

2
aa
3
aba
5
abaab

输出

2 1
3 2 1
5 4 2 1 3

这题比赛时用的暴力，毫无疑问地超时了，赛后发现可以用后缀数组来解决，考了一个定理：
• Let $C_i=mi{n}_{j>iand{s}_j=s_i}j-i$
• The B-Suffix Array is equivalent to the suffix array of $C_1{C}_2...C_n$
拿第三个样例来说，求出它的 $C$ 数组为 $231556$，注意最后的字符后面默认为 $n+1$，求出的后缀数组恰为 $312456$，倒序输出即为答案，但是这题比较麻烦的就是要重写后缀数组的比较函数，因为两个字符之间的位置会很远，以至于超过 ASCILL 码表，所以原来的字符数组必须改为整型数组，重写一下比较函数即可，AC代码如下：

#include
using namespace std;
typedef long long  ll;
const int N=1e5+5;
int n,c[N];
char s[N];
int rk[N],tmp[N],sa[N],k;
void init(){
    k=0;
    memset(rk,0,sizeof(rk));
    memset(tmp,0,sizeof(tmp));
    memset(c,0,sizeof(c));
}
bool cmp(int i,int j){
    if(rk[i]!=rk[j]) return rk[i]<rk[j];
    else{
        int ri,rj;
        if(i+k<=n){
            ri=rk[i+k];
        }
        else{
            ri=-1;
        }
        if(j+k<=n){
            rj=rk[j+k];
        }
        else{
            rj=-1;
        }
        return ri<rj;
    }
}
void get_sa(int sa[]){
    for(int i=0;i<=n;i++){
        sa[i]=i;
        if(i<n){
            rk[i]=c[i];
        }
        else{
            rk[i]=-1;
        }
    }
    for(k=1;k<=n;k*=2){
        sort(sa,sa+n+1,cmp);
        tmp[sa[0]]=0;
        for(int i=1;i<=n;i++){
            tmp[sa[i]]=tmp[sa[i-1]]+(cmp(sa[i-1],sa[i]) ? 1 : 0);
        }
        for(int i=0;i<=n;i++){
            rk[i]=tmp[i];
        }
    }
}
int main(){
    while(~scanf("%d",&n)){
        scanf("%s",s);
        int pos1=n,pos2=n;
        for(int i=n-1;i>=0;i--){
            if(s[i]=='a'){
                c[i]=(n-(pos1==n?pos1:pos1-i))%n;
                pos1=i;
            }else{
                c[i]=(n-(pos2==n?pos2:pos2-i))%n;
                pos2=i;
            }
        }
        get_sa(sa);
        for(int i=1;i<=n;i++) printf("%d ",sa[i]+1);
        printf("\n");
    }
    return 0;
}