Hdu 6194 Shenyang Network Competition -- string (suffix array), hdu -- string

Hdu 6194 Shenyang Network Competition -- string (suffix array), hdu -- string

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Problem DescriptionUncle 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.
 

 

InputThe 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 Σ length (s) ≤ 2 bytes 106.

 

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

 

Sample Input22abcabc3abcabcabcabc

 

Sample Output69. There is a string s. How many substrings exactly appear k times? Idea: suffix array. You can use the suffix array algorithm to know the ranking of each suffix. If a substring occurs exactly k times, then there must be k corresponding suffixes, that is, this substring is the prefix of these k suffix strings, so the ranking of these k suffix strings must be continuous, so we rank from 1 ~ Len (s) starts to fetch k consecutive suffix strings. You can quickly calculate the maximum common prefix length len of the k suffix strings Based on the height [] array, so long as 1 to len prefix substrings, these k strings contain, set the current start k strings to I + k-1, then if the substring length is too short, possible I-1 or I + k this string also contains the corresponding substring, So calculate the I and I-1 string, I + k and I + k-1 maximum common prefix is m, so the length of the substring must be greater than m to ensure that the I-1 and I + k do not contain the corresponding substring, only I ~ The I + k-1 k strings contain corresponding substrings. The Code is as follows:

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <cmath>using namespace std;typedef long long LL;const int N=1e5+5;char s[N];int k;int wa[N],wb[N],wv[N],wss[N];int sa[N],ran[N],height[N];int f[N][20];int cmp(int *r,int a,int b,int l){    return r[a]==r[b]&&r[a+l]==r[b+l];}void da(char *r,int *sa,int n,int m){    int i,j,p,*x=wa,*y=wb,*t;    for(i=0; i<m; i++) wss[i]=0;    for(i=0; i<n; i++) wss[x[i]=(int)r[i]]++;    for(i=1; i<m; i++) wss[i]+=wss[i-1];    for(i=n-1; i>=0; i--) sa[--wss[x[i]]]=i;    for(j=1,p=1; p<n; j*=2,m=p)    {        for(p=0,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<n; i++) wv[i]=x[y[i]];        for(i=0; i<m; i++) wss[i]=0;        for(i=0; i<n; i++) wss[wv[i]]++;        for(i=1; i<m; i++) wss[i]+=wss[i-1];        for(i=n-1; i>=0; i--) sa[--wss[wv[i]]]=y[i];        for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1; i<n; i++)            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;    }    return;}void callheight(char *r,int *sa,int n){    int i,j,k=0;    for(i=1;i<=n;i++)        ran[sa[i]]=i;    for(i=0;i<n;height[ran[i++]]=k)    for(k?k--:0,j=sa[ran[i]-1];r[i+k]==r[j+k];k++);    return ;}void init(int len){    for(int i=1;i<=len;i++) f[i][0]=height[i];    for(int s=1;(1<<s)<=len;s++)    {        int tmp=(1<<s);        for(int i=1;i+tmp-1<=len;i++)        {            f[i][s]=min(f[i][s-1],f[i+tmp/2][s-1]);        }    }}int cal(int l,int r){    int len=log2(r-l+1);    int ans=min(f[l][len],f[r-(1<<len)+1][len]);    return ans;}int main(){    int T; cin>>T;    while(T--)    {       scanf("%d%s",&k,s);       int len=strlen(s);       da(s,sa,len+1,130);       callheight(s,sa,len);       init(len);       int ans=0;       for(int i=1;i+k-1<=len;i++)       {           int j=i+k-1;           int tmp=height[i];           if(j+1<=len) tmp=max(tmp,height[j+1]);           int x;           if(k!=1) { x=cal(i+1,j); }           else x=len-sa[i];           ans+=max(0,x-tmp);       }       printf("%d\n",ans);    }    return 0;}