Here's my 15 minute effort to solve the 2016 Advent of Code Day 1, Part 1 problem.

Here's my 15 minute effort to solve the 2016 Advent of Code Day 1, Part 1 problem. The task is to follow some directions (90° turns and steps on a rectilinear grid), then state the rectilinear distance from the origin.

And here's part 2! This was harder and I had to iterate a little to get the answer to come out.

I had to add a new routine to detect a crossing over the previous route. Of course the easiest way with the in-place solution for part 1 is just to use the last added position as where we're coming from, then iterate through the previous positions and detect when we cross.

I made the passed positions pointer to int so we can update to the crossing point when needed, and still return 1 to break out of reading the routing instructions.

There's obviously a much better and shorter solution lurking in there somewhere, but at least I have something that works.

// http://adventofcode.com/2016/day/1
//
// "C:\Program Files (x86)\Microsoft Visual Studio 14.0\VC\bin\x86_amd64\vcvarsx86_amd64.bat" to setup
// cl day01.c to compile
// day01.exe to run

const char * input = "R4, R5, L5, L5, L3, R2, R1, R1, L5, R5, R2, L1, L3, L4, R3, L1, L1, R2, R3, R3, R1, L3, L5, R3, R1, L1, R1, R2, L1, L4, L5, R4, R2, L192, R5, L2, R53, R1, L5, R73, R5, L5, R186, L3, L2, R1, R3, L3, L3, R1, L4, L2, R3, L5, R4, R3, R1, L1, R5, R2, R1, R1, R1, R3, R2, L1, R5, R1, L5, R2, L2, L4, R3, L1, R4, L5, R4, R3, L5, L3, R4, R2, L5, L5, R2, R3, R5, R4, R2, R1, L1, L5, L2, L3, L4, L5, L4, L5, L1, R3, R4, R5, R3, L5, L4, L3, L1, L4, R2, R5, R5, R4, L2, L4, R3, R1, L2, R5, L5, R1, R1, L1, L5, L5, L2, L1, R5, R2, L4, L1, R4, R3, L3, R1, R5, L1, L4, R2, L3, R5, R3, R1, L3";

#define min(x,y) ((x)<(y) ? (x) : (y))
#define max(x,y) ((x)>(y) ? (x) : (y))

enum {N,E,S,W} direction = N;
int translate_ns[4] = {1,0,-1,0};
int translate_ew[4] = {0,1,0,-1};
int visited[2000];
int visited_count = 1;

int check_add_visited(int* ns, int* ew)
{
    int nsp = visited[(visited_count-1)*2];
    int ewp = visited[(visited_count-1)*2+1];
    // start from 1 so we have a previous entry to check
    int prevns = visited[0];
    int prevew = visited[1];
    for (int i=1;i<visited_count;++i) {
        // get the prev and next ns,ew to check
        int nextns = visited[i*2];
        int nextew = visited[i*2+1];
        if (prevns==nextns && ewp==*ew && min(*ns,nsp)<prevns && max(*ns,nsp)>prevns
                                       && min(prevew,nextew)<*ew && max(prevew,nextew)>*ew) {
            // found a crossing at prevns,ew
            *ns = prevns;
            return 1;
        }
        if (prevew==nextew && nsp==*ns && min(*ew,ewp)<prevew && max(*ew,ewp)>prevew 
                                       && min(prevns,nextns)<*ns && max(prevns,nextns)>*ns ) {
            // found a crossing at ns,prevew
            *ew = prevew;
            return 1;
        }
        prevns = nextns;
        prevew = nextew;
    }
    visited[visited_count*2] = *ns;
    visited[visited_count*2+1] = *ew;
    visited_count++;
    return 0;
}

int main()
{
    int ns = 0;
    int ew = 0;
    int pos = 0;
    char digit = 0;
    int walk = 0;
    visited[0] = visited[1] = 0;
    while (input[pos]!='\0') {
        char turn = input[pos++];
        if (turn=='R') direction = (direction+1) % 4;
        if (turn=='L') direction = (direction+3) % 4; // bwahahah
        walk = 0;
        while ((digit=input[pos++])!='\0' && isdigit(digit)) walk = walk*10+digit-'0';
        ns += walk*translate_ns[direction];
        ew += walk*translate_ew[direction];
        if (check_add_visited(&ns,&ew)) break;
        while (input[pos] && (input[pos]==',' || input[pos]==' ')) pos++;
    }
    printf("Distance to Easter Bunny HQ is: %d",abs(ns)+abs(ew));
}

The output from part 2 is:

C:\Workarea\AdventOfCode\Day01>day01.exe
Distance to Easter Bunny HQ is: 151
C:\Workarea\AdventOfCode\Day01>