Roadmap with tunnels - c

Question

Given a roadmap between a number of cities, with roads between 2 cities containing tunnels, your goal is to find the shortest possible paths between the start city and all the other cities, such that each path contains at least one tunnel. (The problem doesn't always have a solution). Assume the cost of the roads is given. Input - from a file, output - to a file, containing the start city and the path to the rest of the cities.
Now I tried to do this with Dijkstra's algorithm, it solved most of my problem except the part where tunnels are mandatory. Can anyone help me with this? This is my code.
Thanks in advance!
File input:
10
1 2 10
1 4 5
2 3 1
2 4 3
3 5 6
4 2 2
4 3 9
4 5 2
5 1 7
5 3 4
#include <stdio.h>
#define GRAPHSIZE 2048
#define INFINITY GRAPHSIZE*GRAPHSIZE
#define MAX(a, b) ((a > b) ? (a) : (b))
int e; /* The number of nonzero edges in the graph */
int n; /* The number of nodes in the graph */
long dist[GRAPHSIZE][GRAPHSIZE];/* dist[i][j] is the distance between node i and j; or 0 if there is no direct connection */
long d[GRAPHSIZE]; /* d[i] is the length of the shortest path between the source (s) and node i */
int prev[GRAPHSIZE]; /* prev[i] is the node that comes right before i in the shortest path from the source to i*/
void printD() {
int i;
printf("Distances:\n");
for (i = 1; i <= n; ++i)
printf("%10d", i);
printf("\n");
for (i = 1; i <= n; ++i) {
printf("%10ld", d[i]);
}
printf("\n");
}
/*
* Prints the shortest path from the source to dest.
* dijkstra(int) MUST be run at least once BEFORE
* this is called
*/
void printPath(int dest) {
if (prev[dest] != -1)
printPath(prev[dest]);
printf("%d ", dest);
}
void dijkstra(int s) {
int i, k, mini;
int visited[GRAPHSIZE];
for (i = 1; i <= n; ++i) {
d[i] = INFINITY;
prev[i] = -1; /* no path has yet been found to i */
visited[i] = 0; /* the i-th element has not yet been visited */
}
d[s] = 0;
for (k = 1; k <= n; ++k) {
mini = -1;
for (i = 1; i <= n; ++i)
if (!visited[i] && ((mini == -1) || (d[i] < d[mini])))
mini = i;
visited[mini] = 1;
for (i = 1; i <= n; ++i)
if (dist[mini][i])
if (d[mini] + dist[mini][i] < d[i]) {
d[i] = d[mini] + dist[mini][i];
prev[i] = mini;
}
}
}
int main(int argc, char *argv[]) {
int i, j;
int u, v, w;
FILE *fin = fopen("dist.txt", "r");
/* the first line contains e, the number of edges the following e lines
contain 3 numbers: u, v and w signifying that there’s an edge from u to v of weight w*/
fscanf(fin, "%d", &e);
for (i = 0; i < e; ++i)
for (j = 0; j < e; ++j)
dist[i][j] = 0;
n = -1;
for (i = 0; i < e; ++i) {
fscanf(fin, "%d%d%d", &u, &v, &w);
dist[u][v] = w;
n = MAX(u, MAX(v, n));
}
fclose(fin);
dijkstra(1);
printD();
printf("\n");
for (i = 1; i <= n; ++i) {
printf("Path to %d: ", i);
printPath(i);
printf("\n");
}
return 0;
}

Answer 1

Run Dijkstra's algorithm to find all the shortest paths from the starting city to all tunnels.
The run Dijkstra's algorithm again with all the tunnels as starting points to find all the shortest paths to all the other cities. So you'll sort of start in the middle of Dijkstra's algorithm, where you already have a bunch of candidates (all the tunnels) in your priority queue, and all of these will be marked as visited.
It doesn't look like you're using a priority queue (the efficient implementation of Dijkstra's algorithm uses one), but I'm sure you'll manage to figure out how to apply my solution to your code nonetheless.

Answer 2

You can construct a graph like this: have two nodes per city (say C and C' for each city C). For each road, say from C1 to C2, add edges to the graph: C1->C2 and C1'->C2'. For each tunnel, say from C1 to C2, add edges to the graph C1->C2' and C1'->C2'.
The intuition is that to get to a C' node you have to go through at least one tunnel.
Now, to find the shortest path from C1 to C2 using at least one tunnel, simply use Dijkstra to find the shortest path from C1 to C2'. Or to find shortest paths to each city, find all shortest paths from the start city to C' for each city C.