Boyer Moore replace more than one pattern - c

Question

I am working on a string search and replace project. I can only change 1 of the target pattern in the sentence. But I can find both.
Example: just do it. you will do it.
find: do
replace: think
expected---> just think it. you will think it.
what actually happened ---> just do it. you will think it.
How can I replace both of them?
I read the sentence from file input.txt
# include <limits.h>
# include <string.h>
# include <stdio.h>
#include <sys/time.h>
# define NO_OF_CHARS 256
# define MAX 10000
int sum = 0;
int control = 0;
// A utility function to get maximum of two integers
int max (int a, int b) { return (a > b)? a: b; }
// The preprocessing function for Boyer Moore's bad character heuristic
void badCharHeuristic( char *str, int size, int badchar[NO_OF_CHARS]) {
int i;
// Initialize all occurrences as -1
for (i = 0; i < NO_OF_CHARS; i++)
badchar[i] = -1;
// Fill the actual value of last occurrence of a character
for (i = 0; i < size; i++)
badchar[(int) str[i]] = i;
}
/* A pattern searching function that uses Bad Character Heuristic of Boyer Moore Algorithm */
void search( char *txt, char *pat,char temp3[MAX],int k,char*r) {
int m = strlen(pat);
int n = strlen(txt);
char src[MAX],p[MAX],temp[MAX],temp2[MAX],tempP[MAX],out[MAX];
int badchar[NO_OF_CHARS],i,leng,l,count;
char v;
/* Fill the bad character array by calling the preprocessing function badCharHeuristic() for given pattern */
badCharHeuristic(pat, m, badchar);
leng = strlen(pat);
strcpy(tempP,r);
//strcat(tempP,"</mark>");
leng = strlen(pat);
l = strlen(txt);
int s = 0; // s is shift of the pattern with respect to text
while(s <= (n - m)) {
int j = m-1;
/* Keep reducing index j of pattern while characters of pattern and text are matching at this shift s */
while(j >= 0 && pat[j] == txt[s+j]) {
count++;
j--;
}
/* If the pattern is present at current shift, then index j will become -1 after the above loop */
if (j < 0) {
//printf("pattern occurs at shift = %d\n", s);
/* Shift the pattern so that the next character in text
aligns with the last occurrence of it in pattern.
The condition s+m < n is necessary for the case when
pattern occurs at the end of text */
printf("The desired pattern was found starting from %d. line at position %d\n",k,s+1);
strncpy(temp, txt, s);
temp[s] = '\0';
//strcat(temp,"<mark>");
control++;
strcat(temp,tempP);
for(i=0;i<MAX;i++) {
if((s+leng+i)<strlen(txt))
temp2[i] = txt[s+leng+i];
else
temp2[i] = v;
}
strcat(temp,temp2);
strcpy(temp3,temp);
s += (s+m < n)? m-badchar[txt[s+m]] : 1;
}
else
/* Shift the pattern so that the bad character in text
aligns with the last occurrence of it in pattern. The
max function is used to make sure that we get a positive
shift. We may get a negative shift if the last occurrence
of bad character in pattern is on the right side of the
current character. */
s += max(1, j - badchar[txt[s+j]]);
}
sum +=count;
}
/* Driver program to test above funtion */
int main() {
char txt[MAX],p[MAX],r[MAX],temp[MAX],temp2[MAX],tempP[MAX],out[MAX];
int k = 1;
FILE *input = fopen("input.txt","r");
FILE *output = fopen("output.txt","w");
printf("Enter the text in which pattern is to be searched:");
fgets(p, MAX, stdin);
printf("Enter the text in which pattern is to be replaced:");
fgets(r, MAX, stdin);
struct timeval tv1, tv2;
gettimeofday(&tv1, NULL);
p[strlen(p)-1]='\0';
temp[1]='a';
while(!feof(input)){
if(fgets (txt, MAX, input)!=NULL) {
txt[strlen(txt)-1] = '\0';
search(txt, p,temp,k,r);
if(temp[1]!='a') {
fprintf(output,"%s\n",temp);
temp[1]='a';
}
else
fprintf(output,"%s\n",txt);
}
k++;
}
if(control==0) {
printf("\nThe pattern was not found in the given text\n\n");
}
gettimeofday(&tv2, NULL);
printf ("Total time = %f seconds\n", (double) (tv2.tv_usec - tv1.tv_usec) / 1000000 + (double) (tv2.tv_sec - tv1.tv_sec));
fclose(input);
fclose(output);
printf("The number of character comparison: %d\n",sum);
return 0;
}