UVa problem 11364: sample algorithm

To solve UVa problem 11364 (Parking), it would be beneficial to know that the minimal distance Michael has to walk is simply twice the distance from the farthest store to the nearest store. For example, consider the following setup:
A________B______________C_____D
X______________________________

A, B, C and D are the stores while X is Michael’s car. To get to all the stores and return to his car, Michael would have to walk:

A to B
B to C
C to D
D to C
C to B
B to A

In essense, the total distance he would have traveled would be just twice the distance from A to D.

Consider now the following setup:

A________B______________C_____D
__________________X____________

For Michael to get to all stores and return to his car, he would need to walk:

X to B
B to A
A to B
B to X
X to C
C to D
D to C
C to X

Again, analyzing this series of walks, it is simply twice the distance from A to D.

Hence to compute the minimal distance given a series of stores to walk to, simply find the nearest store and the farthest store, get their difference and then multiply the difference by 2.

UVa problem 11364: sample code

001 #include "stdio.h"
002 #include "stdlib.h"
003 
004 int main() {
005   char  line[100]
006       , ctmp[100]
007   ;
008   int   num_tests
009       , num_stores
010       , stores[100]
011       , itmp
012       , max
013       , min
014       , diff
015       , i
016       , j
017       , k
018       , m
019   ;
020 
021   fgets(line, 100, stdin);
022   sscanf(line, "%d\n", &num_tests);
023 
024   for (i = 0; i < num_tests; i += 1) {
025     fgets(line, 100, stdin);
026     sscanf(line, "%d\n", &num_stores);
027 
028     fgets(line, 100, stdin);
029     for (j = 0, m = 0; 1; j += 1) {
030       if (line[j] == '\n' || line[j] == 0) {
031         break;
032       }
033 
034       if (line[j] == ' ' || line[j] == '\t') {
035         continue;
036       }
037 
038       for (k = 0; 1 ; j += 1, k += 1) {
039         if (line[j] == ' ' || line[j] == '\t' || line[j] == '\n' || line[j] == 0) {
040           break;
041         }
042         ctmp[k] = line[j];
043       }
044       ctmp[k] = 0;
045 
046       stores[m] = atoi(ctmp);
047       m += 1;
048     }
049     
050     max = 0;
051     min = 99;
052     for (j = 0; j < num_stores; j += 1) {
053       if (stores[j] > max) {
054         max = stores[j];
055       }
056 
057       if (stores[j] < min) {
058         min = stores[j];
059       }
060     }
061     diff = max - min;
062     diff *= 2;
063 
064     printf("%d\n", diff);
065   }
066 
067   return 0;
068 }