name.c

/*    name.c
    Corporate name changing program -- string example

    by:Steven Skiena
    begun: January 28, 2002
    revised: July 3, 2002
*/


/*
Copyright 2003 by Steven S. Skiena; all rights reserved. 

Permission is granted for use in non-commerical applications
provided this copyright notice remains intact and unchanged.

This program appears in my book:

"Programming Challenges: The Programming Contest Training Manual"
by Steven Skiena and Miguel Revilla, Springer-Verlag, New York 2003.

See our website www.programming-challenges.com for additional information.

This book can be ordered from Amazon.com at

http://www.amazon.com/exec/obidos/ASIN/0387001638/thealgorithmrepo/

*/

#include <stdio.h>
#include <string.h>

#define MAXLEN      1001   /* longest possible string */
#define MAXCHANGES  101    /* maximum number of name changes */

typedef char string[MAXLEN];

string mergers[MAXCHANGES][2];   /* store before/after corporate names */
int nmergers;                    /* the number of different name changes */

void read_quoted_string(char *s) {
    int i = 0;      /* counter */
    char c;         /* latest character */

    while ((c = getchar()) != '\"')
        ;

    while ((c = getchar()) != '\"') {
        s[i] = c;
        i = i + 1;
    }

    s[i] = '\0';
}

void read_changes(void) {
    int i;    /* counter */

    scanf("%d\n", &nmergers);
    for (i = 0; i<nmergers; i++) {
        read_quoted_string((char*)&(mergers[i][0]));
        read_quoted_string((char*)&(mergers[i][1]));
    }
}

/*    Replace the substring of length xlen starting at position pos in
      string s with the contents of string y.
*/

void replace_x_with_y(char *s, int pos, int xlen, char *y) {
    int i;            /* counter */
    int slen,ylen;    /* lengths of relevant strings */

    slen = strlen(s);
    ylen = strlen(y);

    if (xlen >= ylen) {
        for (i = (pos+xlen); i <= slen; i++) {
            s[i+(ylen-xlen)] = s[i];
        }
    } else {
        for (i = slen; i >= (pos+xlen); i--) {
            s[i+(ylen-xlen)] = s[i];
        }
    }

    for (i = 0; i < ylen; i++) {
        s[pos+i] = y[i];
    }
}

/*    Return the position of the first occurrence of the pattern p
      in the text t, and -1 if it does not occur.
*/
/* [[[ findmatch_cut */
int findmatch(char *p, char *t) {
    int i, j;          /* counters */
    int plen, tlen;    /* string lengths */

    plen = strlen(p);
    tlen = strlen(t);

    for (i = 0; i <= (tlen-plen); i = i + 1) {
        j=0;
        while ((j < plen) && (t[i + j] == p[j])) {
            j = j + 1;
        }
        if (j == plen) {
            return(i);
        }
    }
    return(-1);
}
/* ]]] */

int main(void) {
    string s;    /* input string */
    char c;      /* input character */
    int nlines;  /* number of lines in text */
    int i,j;     /* counters */
    int pos;     /* position of pattern in string */

    read_changes();
    scanf("%d\n", &nlines);

    for (i = 1; i <= nlines; i = i + 1) {
        /* read text line */
        j = 0;
        while ((c = getchar()) != '\n') {
            s[j] = c;
            j = j + 1;
        }
        s[j] = '\0';

        for (j = 0; j < nmergers; j = j + 1) {
            while ((pos=findmatch(mergers[j][0], s)) != -1) {
                replace_x_with_y(s, pos, strlen(mergers[j][0]), mergers[j][1]);
            }
        }
        printf("%s\n", s);
    }
    return 0;
}