Hdoj title 3518 boring counting (suffix array, number of sub-strings with a minimum of 2 repetitions)

Boring countingTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 2253 Accepted Submission (s): 924


Problem Description035 now faced a tough problem,his 中文版 teacher gives him a String,which consists with n lower case l Etter,he must figure out how to many substrings appear at least twice,moreover,such apearances can not overlap all other.
Take the AAAA as an example. " A "apears four times," AA "apears-without overlaping.however,aaa can ' t apear more than one time without overlapin G.since we can get "AAA" from [0-2] (the position of string begins with 0) and [1-3]. But the interval [0-2] and [1-3] overlaps each other. So "AAA" can does not take into account. Therefore,the answer is 2 ("a", and "AA").

Inputthe input data consist with several test cases. The input ends with a line "#". Each test case contain a string consists with lower letter,the length n won ' t exceed (n <= 1000).
Outputfor each test case output a integer Ans,which represent the answer for the test case.you ' d better use Int64 to Avoi D unnecessary trouble.
Sample Input

aaaaababcabbaaaaaa#

Sample Output

233

Source2010 acm-icpc multi-university Training Contest (9)--host by Hnu
Recommendzhengfeng | We have carefully selected several similar problems for you:3517 3520 3519 3521 3522

problem:3518 (boring counting)     Judge status:acceptedrunid:14564325    language:c++    Author:lwj1994code Render status:rendered by hdoj C + + Code Render Version 0.01 Beta

AC Code

#include <stdio.h> #include <string.h> #include <algorithm> #include <iostream&gt         ;      using namespace Std;     int s[2002];   Char str[2002];      int sa[2002],t1[2002],t2[2002],c[2002];      int Rank[2002],height[2002],ans;          void Build_sa (int s[],int n,int m) {int i,j,p,*x=t1,*y=t2;          for (i=0;i<m;i++) c[i]=0;          for (i=0;i<n;i++) c[x[i]=s[i]]++;          for (i=1;i<m;i++) c[i]+=c[i-1];          for (i=n-1;i>=0;i--) sa[--c[x[i]]]=i;              for (j=1;j<=n;j<<=1) {p=0;              for (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<m;i++) c[i]=0;              for (i=0;i<n;i++) c[x[y[i]]]++;       for (i=1;i<m;i++) c[i]+=c[i-1];       for (i=n-1;i>=0;i--) sa[--c[x[y[i]]]]=y[i];              Swap (x, y);              P=1;              x[sa[0]]=0;              for (i=1;i<n;i++) x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+j]==y[sa[i]+j]?p-1:p++;              if (p>=n) break;          M=p;          }} void GetHeight (int s[],int n) {int i,j,k=0;          for (i=0;i<=n;i++) rank[sa[i]]=i;              for (i=0;i<n;i++) {if (k) k--;              J=SA[RANK[I]-1];              while (S[i+k]==s[j+k]) k++;          Height[rank[i]]=k; }}int judge (int n,int len) {int maxn=sa[0],minn=sa[0],ans=0;int i,j;for (i=1;i<=n;i++) {if (Height[i]<len) {if (max N-minn>=len) ans++;maxn=minn=sa[i];} Else{if (Maxn<sa[i]) maxn=sa[i];if (Minn>sa[i]) minn=sa[i];}} if (Maxn-minn>=len) Ans++;return ans;} int main () {int n;while (scanf ("%s", str)!=eof) {int i;if (strcmp (StR, "#") ==0) break;int Len=strlen (str); for (i=0;i<len;i++) s[i]=str[i]-' a ' +1;s[len]=0;build_sa (s,len+1,30); GetHeight (s,len); int l=0,r=len;ans=0;for (i=1;i<=len/2;i++) {Ans+=judge (len,i);} printf ("%d\n", ans);}}

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Hdoj title 3518 boring counting (suffix array, number of sub-strings with a minimum of 2 repetitions)