HDU 5769 Substring (suffix array)

Substring

Time limit:2000/1000 MS (java/others) Memory limit:65536/65536 K (java/others)
Total submission (s): 1570 Accepted Submission (s): 618

Problem Description?? Is practicing his program skill, and now he's given a string, he has to calculate the total number of its distinct substr Ings.
But?? Thinks that's too easy, he wants to make this problem more interesting.
?? Likes a character x very much, so he wants to know the number of distinct substrings which contains at least one X.
However,?? is unable to solve it, please help him.

Inputthe first line of the input gives the number of test cases T; T test Cases follow.
Each test case is consist of 2 lines:
First line was a character X, and second line is a string S.
X is a lowercase letter, and S contains lowercase letters (' a '-' Z ') only.

T<=30
1<=| S|<=10^5
The Sum of | s| In all the test cases are no more than 700,000.

Outputfor each test case, output one line containing ' case #x: Y ' (without quotes), where x is the ' test Case number (Startin G from 1) and Y-is the answer.

Sample INPUT2 a ABC b BBB

Sample outputcase #1:3 case #2:3 HintIn first case, all distinct substrings containing at least one a:a, AB, ABC. In second case, all distinct substrings containing at least one b:b, BB, BBB.

Authorfzu Analysis: The topic of a particular letter of the non-repeating substring, we know that the non-repeating substring is the basic use of the suffix array, you can first find the total non-repeating substring. Then the non-repeating substring containing the letter is not included, and the last two are subtracted, that is, when the non-repeating substring that does not contain the letter is evaluated, You can first find the total number of substrings that do not contain the letter. When you go to heavy, subtract only the height value that does not contain a specific letter. In this case, we need to record each letter with an array, the position of the nearest specific letter on the right. The code is as follows:

#include <cstdio>#include<iostream>#include<algorithm>#include<cstring>typedefLong Longll;using namespacestd;Const intmaxn=200010;intR[MAXN];intWA[MAXN],WB[MAXN],WV[MAXN],WS[MAXN];CharSTR[MAXN];intcmpint*r,intAintBintl) {returnr[a]==r[b]&&r[a+l]==r[b+l];}voidDaConst CharR[],intSa[],intNintM//N is len+1,m generally larger than the largest number in the array{      inti,j,p,*x=wa,*y=wb,*T;  for(i=0; i<m; i++) ws[i]=0;  for(i=0; i<n; i++) ws[x[i]=r[i]]++;  for(i=1; i<m; i++) ws[i]+=ws[i-1];  for(i=n-1; i>=0; i--) sa[--ws[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]&GT;=J) y[p++]=sa[i]-J;  for(i=0; i<n; i++) wv[i]=X[y[i]];  for(i=0; i<m; i++) ws[i]=0;  for(i=0; i<n; i++) ws[wv[i]]++;  for(i=1; i<m; i++) ws[i]+=ws[i-1];  for(i=n-1; i>=0; i--) sa[--ws[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)? p1:p + +; }      return;}intSA[MAXN],RANK[MAXN],HEIGHT[MAXN];//The SA is found in the position of the string by the suffix rank, where rank is the suffix rank based on the position, the inverse of which is the minimum value of the SA array in the template is 1. voidCalheight (Const Char*r,int*sa,intN) {      inti,j,k=0;  for(i=1; i<=n; i++) rank[sa[i]]=i;  for(i=0; i<n; height[rank[i++]]=k) for(k?k--:0, j=sa[rank[i]-1]; R[I+K]==R[J+K]; k++);}intMain () {intT,len,case=0;; Charch; scanf ("%d",&t);  while(t--) { case++; ll Ans2=0; Memset (R,0,sizeof(R)); scanf ("%c",&ch); scanf ("%s", str); Len=strlen (str); intRr=Len;  for(inti=len-1; i>=0; i--)      {         if(str[i]!=ch) {R[i]=RR; }         Else{ll h= (rr-i-1); RR=i; R[i]=RR; Ans2+=h* (H +1)/2; }} ans2+=rr* (rr+1)/2; //cout<<ans2<<endl;Da (str,sa,len+1, the);       Calheight (Str,sa,len);  for(intI=2; i<=len;i++) {Ans2-=min (height[i],r[sa[i]]-Sa[i]); } ll ans=(Long Long) len* (len+1)/2;  for(intI=2; i<=len;i++) {ans-=Height[i]; }    //cout<<ans<< "" <<ans2<<endl;printf"Case #%d:%lld\n", case,ans-ans2); }return 0;}

HDU 5769 Substring (suffix array)