hihocoder-1415 suffix Array Three • The longest common substring of a repeating melody of 32 strings

The S1,S2 stitching, to find height. The adjacent height determines whether the starting point of the left and right string is in two strings, and the height and s1.length ()-sa[i-1] Take min.

#include <iostream>#include<cstring>#include<string>#include<queue>#include<vector>#include<map>#include<Set>#include<stack>#include<cmath>#include<cstdio>#include<algorithm>#include<iomanip>#include<stdlib.h>#include<time.h>#defineLL Long Longusing namespacestd;ConstLL mod =100001;ConstLL N =300010;ints1l, s2l;classsf{//N: Array size Public:    intX[n], y[n], c[n]; intHeight[n],str[n], Sa[n], rank[n];//height array starting from 2    intSlen; intm=1050;//Character Set processing size (passing in if not a number, need to do displacement conversion)    BOOLcmpint* R,intAintBintl) {returnR[a] = = R[b] && r[a + l] = = R[b +L]; }    voidSuffix (intN) {++N; intI, J, p;  for(i =0; I < m; ++i) C[i] =0;  for(i =0; I < n; ++i) C[x[i] = str[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]]= CMP (y, Sa[i-1], Sa[i], j)? P-1: p++; }            if(P >= N) Break; M=p; }        intK =0; N--;  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(Str[i + K] = = Str[j + K]) + +K; Height[rank[i]]=K; //cout << k << Endl;        }    }    voidInitstringS//no matter what parameter, pass in with reference{Slen=s.length ();  for(inti =0; i < Slen; i++) Str[i]=s[i]-'a'+2;//if it is a character, it is mapped to a sequence starting from 1Str[slen] =1;//0 as Terminator to prevent cross-borderSuffix (Slen); } LL Solve () {intAns =0;  for(inti =2; I <= Slen; i++)        {            intSA1 = Sa[i-1],sa2=Sa[i]; if(SA1 >sa2) Swap (SA1, SA2); if(SA1 >= s1l | | SA2 < S1L)Continue; Ans= Max (ans, min (s1l-SA1, height[i]); }        returnans; }}SF; LL dp[ *][2]; LL N;intMain () {Cin.sync_with_stdio (false); strings1,s2;  while(Cin >> S1>>S2) {s1l=s1.length (); S2L=s2.length (); Sf.init (S1+S2); cout<< sf.solve () <<Endl; }    return 0;}

hihocoder-1415 suffix Array Three • The longest common substring of a repeating melody of 32 strings