Hdu 6194 Shenyang Online race--string string string (suffix array)

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problem DescriptionUncle Mao is a wonderful acmer. One day he met a easy problem, but Uncle Mao is so lazy this he left the problem. 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 the find out how many substrings which is important strings.

InputThe first line contains an integerT(T≤ ) implying the number of test cases.
For each test case, there is lines:
The first line contains an integerk(k≥1 ) which is described above;
The second line contain a strings(lengtH(s)≤5 ).
It ' s guaranteed that∑ l en gt h ( S) ≤ 2∗ 10 6 .

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

Sample Input22abcabc3abcabcabcabc

Sample Output69 Test Instructions: There is a string s, how many of the substrings that happen to have exactly k times? Idea: suffix array, through the suffix array algorithm can know the rank of each suffix, if there is a substring happens to happen K, then there must be k corresponding suffix that this substring is the prefix of this k suffix string, then this k suffix string ranking must be continuous, so we ranked from 1~len (s) sequentially Take a continuous k suffix string, can be based on the height[] array to quickly calculate the current K suffix string of the maximum common prefix length len, then a length of 1 to Len prefix substring, this k string contains, set the current start K strings for I to i+k-1, then if the substring length is too short, may i-1 or i+k This string also contains the corresponding substring, so calculate I and i-1 string, i+k and i+k-1 the maximum common prefix is m, then the substring length must be greater than m to ensure that i-1 and i+k does not contain the corresponding substring, only i~i+k-1 this k string contains the corresponding substring. The code is as follows:

#include <iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>using namespaceStd;typedefLong LongLL;Const intn=1e5+5;CharS[n];intK;intWa[n],wb[n],wv[n],wss[n];intSa[n],ran[n],height[n];intf[n][ -];intcmpint*r,intAintBintl) {    returnr[a]==r[b]&&r[a+l]==r[b+l];}voidDaChar*r,int*sa,intNintm) {    inti,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]&GT;=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)? p1:p + +; }    return;}voidCallheight (Char*r,int*sa,intN) {    inti,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 ;}voidInitintLen) {     for(intI=1; i<=len;i++) f[i][0]=Height[i];  for(ints=1;(1<<s) <=len;s++)    {        intTmp= (1<<s);  for(intI=1; i+tmp-1<=len;i++) {F[i][s]=min (f[i][s-1],f[i+tmp/2][s-1]); }    }}intCalintLintR) {    intLEN=LOG2 (r-l+1); intAns=min (f[l][len],f[r-(1<<len) +1][len]); returnans;}intMain () {intT Cin>>u;  while(t--) {scanf ("%d%s",&k,s); intlen=strlen (s); Da (S,sa,len+1, the);       Callheight (S,sa,len);       Init (len); intans=0;  for(intI=1; i+k-1<=len;i++)       {           intj=i+k-1; inttmp=Height[i]; if(j+1<=len) Tmp=max (tmp,height[j+1]); intx; if(k!=1) {x=cal (i+1, J); }           Elsex=len-Sa[i]; Ans+=max (0, X-tmp); } printf ("%d\n", ans); }    return 0;}

Hdu 6194 Shenyang Online race--string string string (suffix array)