HDU 6194 后缀数组

string string string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1850    Accepted Submission(s): 546

Problem Description

Uncle Mao is a wonderful ACMER. One day he met an easy problem, but Uncle Mao was so lazy that he left the problem to you. I hope you can give him a solution.
Given a string s, we define a substring that happens exactly k times as an important string, and you need to find out how many substrings which are important strings.

 

Input

The first line contains an integer T (T≤100) implying the number of test cases.
For each test case, there are two lines:
the first line contains an integer k (k≥1) which is described above;
the second line contain a string s (length(s)≤105).
It's guaranteed that ∑length(s)≤2∗106.

 

Output

For each test case, print the number of the important substrings in a line.

 

Sample Input

2

2

abcabc

3

abcabcabcabc

 

Sample Output

6
9

 

Source

2017 ACM/ICPC Asia Regional Shenyang Online

 

题意：

给一个字符串求其中恰好出现k次的子串的个数

代码：

//我们考虑每一个子串对结果的贡献，每个子串都是某个后缀的前缀，我们得到heigh数组之后可以维护一个长度是k的窗口，这个窗口中的后缀的
//lcp必然出现至少k次，然后再依次减去窗口左右端分别延长一位的lcp(即减去出现次数多于k次的lcp)，这样他们中间的公共部分减了两次再加
//上就行了。注意临界端点的处理！
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN=;
int sa[MAXN+],ra[MAXN+],he[MAXN+],xx[MAXN+],yy[MAXN+],buc[MAXN+];
char s[MAXN+];
int len,m,f[MAXN+][];
void get_suf()
{
    int *x=xx,*y=yy;
    for(int i=;i<m;i++) buc[i]=;
    for(int i=;i<len;i++) buc[x[i]=s[i]]++;
    for(int i=;i<m;i++) buc[i]+=buc[i-];
    for(int i=len-;i>=;i--) sa[--buc[x[i]]]=i;
    for(int k=;k<=len;k<<=){
        int p=;
        for(int i=len-;i>=len-k;i--) y[p++]=i;
        for(int i=;i<len;i++) if(sa[i]>=k) y[p++]=sa[i]-k;
        for(int i=;i<m;i++) buc[i]=;
        for(int i=;i<len;i++) buc[x[y[i]]]++;
        for(int i=;i<m;i++) buc[i]+=buc[i-];
        for(int i=len-;i>=;i--) sa[--buc[x[y[i]]]]=y[i];
        swap(x,y);
        p=;x[sa[]]=;
        for(int i=;i<len;i++){
            if(y[sa[i-]]==y[sa[i]]&&y[sa[i-]+k]==y[sa[i]+k])
                x[sa[i]]=p-;
            else x[sa[i]]=p++;
        }
        if(p>=len) break;
        m=p;
    }
    for(int i=;i<len;i++) ra[sa[i]]=i;
    int k=;
    for(int i=;i<len;i++){
        if(ra[i]==) { he[]=; continue; }
        if(k) k--;
        int j=sa[ra[i]-];
        while(s[i+k]==s[j+k]&&i+k<len&&j+k<len) k++;
        he[ra[i]]=k;
    }
}
void rmq1()
{
    for(int i=;i<=len;i++)
        f[i][]=he[i];
    for(int j=;(<<j)<=len;j++){
        for(int i=;i+(<<j)-<=len;i++)
            f[i][j]=min(f[i][j-],f[i+(<<(j-))][j-]);
    }
}
int query(int l,int k)
{
    if(l<||l+k->=len) return ;
    if(k==) return len-sa[l];
    int j=,r=l+k-;l++;
    while(<<(j+)<=r-l+) j++;
    return min(f[l][j],f[r-(<<j)+][j]);
}
int main()
{
    //freopen("in.txt","r",stdin);
    int t,k;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&k);
        scanf("%s",s);
        len=strlen(s);
        m=;
        get_suf();
        he[]=he[len]=;
        rmq1();
        int ans=;
        for(int i=;i+k-<len;i++)
            ans+=query(i,k)-query(i-,k+)-query(i,k+)+query(i-,k+);
        printf("%d\n",ans);
    }
    return ;
}