#include <stream.h>
#include <string.h>
inline int max(int a, int b) {return a > b ? a : b;}

/* tools for diagnostics and illustration */
int count; /* count of number of character comparisons used */
int count_init();
int count_n_show(char* a, char* p, int i, int j);

const MaxPatternLength = 100;

/* tools for the two string search heuristics used in the Boyer-Moore alg. */
int skip[256];
void initskip(char* p);

int shift[MaxPatternLength];
void initshift(char* p);

int bmsearch(char* p, char* a){
/*
Boyer-Moore string search.
Returns index of first occurrence of string p in a.
Returns length of a if there is no occurrence of p in a.
*/
  int M = strlen(p), 
      M1 = M - 1, 
      N = strlen(a),
      right_end = M1, /* position in a */
      i; /* offset from right in a and p */
  initskip(p);
  initshift(p);
  count_init();
  while (right_end < N){
    /* 1. Get the offset from right of the first non-match */
    for (i = 0; i < M && count_n_show(a, p, right_end - i, M1 - i)
	              && a[right_end - i] == p[M1 - i]; i++); 
    if (i == M) return right_end - M1; /* because a match has been found */

    /* 2. Figure skip right that would align current character in a 
	  with the rightmost occurrence (if any) of that character in p */
    int sk = skip[a[right_end - i]];

    /* 3. Figure shift right that would align the current matched initial
	  segment with the next such segment in p */
    int sh = shift[i];

    /* 4. Perform the largest shift to the right determined by 2. or 3. above */
    right_end = max(right_end - i + sk, right_end + sh);
  }
  return N; /* because no match was found */
}

void initskip(char* p){
/*
initializes the skip array. skip[c] = j if the rightmost occurrence of char c
in p is in location j.  skip[c] = length of p if there is no occurrence of c.
*/
  int M = strlen(p); int M1 = M - 1;
  for (int i = 0; i < 256; i++) skip[i] = M;
  for (i = 0; p[i]; i++) skip[p[i]] = M1 - i;
}


void initshift(char* p){
/*
initializes the shift array for the "pattern" string p.
*/
  int i, j, M = strlen(p); int M1 = M - 1;
  int len[MaxPatternLength];
  /* 1. Set len[i] = number of characters right ended at i which match the
	characters right ended at p[M1]. */
  for (i = 1; i < M; i++){ 
    for (j = 0; j < M && p[M1 - j] == p[M1 - i - j]; j++);
    len[i] = j;
  }
  /* 2. If p[M1] does not occur again in p, then this initialization
        would be the proper shift */
  shift[0] = 1;
  for (i = 1; i < M; i++) shift[i] = M;
  /* 3. Fix up by shifting to the rightmost occurrence of the matched chars */
  for (i = M1; i > 0; i--) shift[len[i]] = i;
  /* 4. Fix up by considering matches that would run off the end of the 
	pattern */
  int ended = 0;
  for (i = 0; i < M; i++){
    if (len[i] == M1 - i) ended = i;
    if (ended) shift[i] = ended;
  }
}

int main(){
/* 
tests bmsearch 
*/
  int index;
  char* p = "bad";
  char* a = "Tad had had \"had\" where Gladis had had \"had had.\"  Too bad Tad
  hadn't had \"had had.\"\n";
  cout << "\nGiven pattern \n\"" << p << "\" and text \n\"" << a << "\"\n";
  index = bmsearch(p, a);
  cout << "bmsearch returns " << index 
       << ", M is " << strlen(p) << ", N is " << strlen(a) << "\n";

  a = "hello, world";
  cout << " Given pattern \n\"" << p << "\" and text \n\"" << a << "\"\n";
  index = bmsearch(p, a);
  cout << "bmsearch returns " << index 
       << ", M is " << strlen(p) << ", N is " << strlen(a) << "\n";

  a = "Goodbye, cruel world";
  cout << " Given pattern \n\"" << p << "\" and text \n\"" << a << "\"\n";
  index = bmsearch(p, a);
  cout << "bmsearch returns " << index
       << ", M is " << strlen(p) << ", N is " << strlen(a) << "\n";
}

int count_init() {count = 0;} /* initialize for diagnostics of search */ 
int count_n_show(char* a, char* p, int i, int j){ /* diagnostics of search */ 
  count++; 
  cout << "comparison of a[" << i << "] = " << a[i] ;
  cout << " with p[" << j << "] = " << p[j] << "\n";
  1; /* return true */
}