POJ 3261 overlapping K-times longest repeating substring "suffix array"

This is an example.

Given a string, the longest repeating substring that appears at least k times, which can overlap.
Algorithm Analysis:
This problem is similar to the previous question, is the first two answers, and then divided into groups of suffixes. No
The same is, here to determine whether there is a group suffix number is not less than K. If there is, then there is
K the same substring satisfies a condition, otherwise it does not exist. The time complexity of this procedure is O (NLOGN).

Source Code:

//#pragma COMMENT (linker, "/stack:16777216")//For C + + Compiler#include <stdio.h>#include<iostream>#include<fstream>#include<cstring>#include<cmath>#include<stack>#include<string>#include<map>#include<Set>#include<list>#include<queue>#include<vector>#include<algorithm>#defineMax (b) ((a) > (b))? (a): (b))#defineMin (b) ((a) < (a))? (a): (b))#defineAbs (x) (((x) > 0)? (x): (-(x)))#defineMOD 1000000007#definePi ACOs (-1.0)using namespaceStd;typedefLong Longll; typedef unsignedLong Longull; typedef unsignedint        UINT; typedef unsignedCharUchar; template<classT> InlinevoidCheckmin (T &a,t b) {if(a>b) a=b;} Template<classT> InlinevoidCheckmax (T &a,t b) {if(a<b) a=b;}Const DoubleEPS = 1e-7      ;Const intN =1              ;Const intM =200000         ;Constll P =10000000097ll;Const intINF =0x3f3f3f3f   ;intA[m], sa[m], h[m], rank[m];voidRadixint*STR,int*a,int*b,intNintm) {    Static intCount[m]; inti; memset (Count,0,sizeof(count));  for(i =0; I < n; ++i) + +Count[str[a[i]];  for(i =1; I <= m; ++i) Count[i] + = count[i-1];  for(i = n-1; I >=0; -i) b[--count[str[a[i] []] =a[i];}voidSuffix_array (int*STR,int*sa,intNintm) {    Static intA[m], b[m]; intI, J;  for(i =0; I < n; ++i) Rank[i] =i;    Radix (str, rank, SA, N, M); rank[sa[0]] =0;  for(i =1; I < n; ++i) Rank[sa[i] = rank[sa[i-1]] + (str[sa[i]]! = str[sa[i-1]]);  for(i =0;1<<i < n; ++i) {         for(j =0; J < N; ++j) {A[j]= Rank[j] +1; B[J]= j + (1<< i) >= n?0: Rank[j + (1<< i)] +1; SA[J]=J;        } radix (b, SA, rank, n, N);        Radix (A, rank, SA, n, N); rank[sa[0]] =0;  for(j =1; J < N; ++j) {Rank[sa[j]]= Rank[sa[j-1]] + (A[SA[J-1]]! = A[sa[j] | | B[sa[j-1]] !=B[sa[j]]); }    }}voidCalc_height (int*STR,int*sa,int*h,intN) {    intI, k =0; h[0] =0;  for(i =0; I < n; ++i) {k= k = =0?0K1; if(Rank[i]! =0){             while(Str[i + K] = = Str[sa[rank[i]-1] +K]) {                ++K; }} H[rank[i]]=K; }}voidSOLVE_DUPLICATE_SUBSTR (intN) {    intI, J, pos, ans =0;  for(i =0; I < n; ++i) {        if(H[rank[i]) >ans) {ans=H[rank[i]]; POS=i; }    }     for(i = pos; I < pos + ans; + +)i) {printf ("%c", A[i]); } printf ("\ n");}voidSLOVE_UPDATE_DUPLICATE_SUBSTR (intNintk) {    intI, J; intLow =1, high =N; intAns =0, POS1 =0, Pos2 =0;  while(Low <=High ) {        intMid = (low + high)/2; BOOLFlag =false;  for(i =0; I < n; ++i) {            if(H[i] >=mid) {                ++ans; if(ans >= k) flag =true; }            ElseAns =1; }        if(flag) Low = mid +1; ElseHigh = mid-1; } cout<< High <<Endl;}intMain () {intI, J, T, N, M, K;  while(Cin >> N >>k) {         for(i =0; I < n; ++i) Cin >>A[i]; Suffix_array (A, SA, N, the);        Calc_height (A, SA, H, N);    Slove_update_duplicate_substr (n, k); }    return 0;}

POJ 3261 overlapping K-times longest repeating substring "suffix array"