Let's make a map of dungeon

Question

Long time ago, when I was spending more time playing RPG, one of the issues what some players had, was to track party movement and to draw proper map. So I come with idea, to check how you guys come deal with this issue.

Task is to write a function, which takes as input parameter list of directions (passed as structure of your choice) ^v<>, display map of dungeon. In example for input: >>>vvv<<<^^^ output will be:

+----+               +----+
|    |               |>>>v|
| ++ |               |^++v|
| ++ |  because      |^++v|
|    |               |^<<<|
+----+               +----+

Testcases

>>>>>>vvvvv<<<<^^^>>v



 +-------+
 |       |
 +-+---+ |
   |   | |
   | + | |
   | +-+ |
   |     |
   +-----+


^^^^^vvv<<<^^vv>>>>>>>>vv>><<^^^^v>>>>v^<^

          +-+ 
       +-+| | +-++-+
       | || | | ++ ++
       | ++ +-+     |
       |        +-+ |
       +--+ +-+ +-+-+
          | | |   |
          +-+ +---+

As it is golf code challenge, shortest code wins.

Happy golfing.

EDIT Sorry for late edit, I didn't have much time recently.

Map is generated based on movement. It should contain only corridors which were visited during walk. Hence creating one big room won't be a valid answer.

There are three valid symbols in the map:

• | vertical wall
• | horizontal wall
• + intersection of vertical and horizontal wall.

Maximum length of the path is 255 character (but if you can, do not limit yourself).

More test cases:

 ><><><><

 +--+
 |  |
 +--+

 >^<v

 +--+
 |  |
 |  |
 +--+

I hope now everything is clear.

Answer 1

Javascript (ES6), 261 254 243 bytes

s=>{v=Array(w=32);p=526;b=n=>v[n>>5]&(1<<(n&31));B=(n,i)=>b(p+n)|b(p-n)?i:0;[...0+s].map(c=>v[(p+=[0,1,-1,w,-w]['0><v^'.indexOf(c)])>>5]|=1<<(p&31));for(z='',p=0;p<w*w;z+=' |-+'[b(p)?0:B(1,1)|B(w,2)||B(31,3)|B(33,3)],p++%w||(z+=`
`));return z}

JSFiddle

Answer 2

C, 246 bytes

n,y,*g,*p;main(c,v)char**v;{for(n=c*2+3,g=calloc(n*n,8),p=g+(n+1)*(c+1);c--;y=*v[c],p-=(y>70?n:1)*(y%5%4?-1:1))*p=1;for(y=n-1;--y;)for(c=n-1;c--;putchar(c?" |-++|-+"[*p?0:p[1]|p[-1]|(p[n]|p[-n])*2|(p[1+n]|p[1-n]|p[n-1]|p[-n-1])*4]:10))p=g+y*n+c;}

Takes input as separate characters, for example:

./mapper '>' '>' '>' 'v' 'v' 'v' '<' '<' '<' '^' '^' '^'

Or more conveniently, this (intentionally unquoted!) format:

./mapper $(echo '>>>vvv<<<^^^' | fold -w1)

Or for (pretty inefficient) random input:

./mapper $(LC_CTYPE=C tr -dc '0-3' < /dev/urandom | tr '0123' '<>^v' | head -c 10 | fold -w1)

And finally, using awk to crop the result means we can get much bigger:

./mapper $(LC_CTYPE=C tr -dc '0-3' < /dev/urandom | tr '0123' '<>^v' | head -c 500 | fold -w1) | awk '/[^ ]/{l=match($0,"\\+");if(l&&(l<L||!L))L=l;v[i++]=$0}END{for(;j<i;){l=v[j++];print substr(l,L,match(l," *$")-L)}}'

---

 +-----+
 |     |
 |     +-+  +-----+
 |       |+-+     |
++       +|      ++
|         |      |
|         +      +-+
+--+               |
 +-+               +--+
 |     ++             |
 +-+  +-++           +-+
   ++   |+--+          |
    +-+ | |        ++  |
     ++ | ++   + +--+ ++
     |  |  +---|      |
     |  |     ++  +---+
     ++ |-+   |   |
      | | +--+-+  |
      | +    +    |
      |        +--+
      +--------+

---

Outputs to stdout. Uses the fact that padding is allowed around the map (produces a map which has edge length 2*n+1, putting the final position in the middle).

Breakdown

This works by looping over the arguments in reverse and moving in reverse. It actually uses the fact that arg 0 is the program name; it doesn't matter what the name is, but it allows us to visit both the initial and last cell (as well as all cells in-between) without needing special handling.

n,                                  // Size of internal grid
y,                                  // Loop counter / current character
*g,                                 // Internal grid memory
*p;                                 // Current working pointer
main(c,v)char**v;{                  // K&R style function declaration
    for(                            // Step 1: identify visited cells
        n=c*2+3,                    //  Set output grid size
        g=calloc(n*n,8),            //  Allocate map storage space
        p=g+(n+1)*(c+1);            //  Start at centre
        c--;                        //  Loop over directions in reverse
        y=*v[c],                    //  Get current direction
        p-=(y>70?n:1)*(y%5%4?-1:1)  //  Move in reverse
    )*p=1;                          //  Mark cell visited
    for(y=n-1;--y;)                 // For each row (except edges)
        for(c=n-1;c--;              //   For each column (except edges, +1 for \n)
            putchar(c?" |-++|-+"[   //   Print wall using lookup table
                *p?0:p[1]|p[-1]|(p[n]|p[-n])*2|(p[1+n]|p[1-n]|p[n-1]|p[-n-1])*4
            ]:10)                   //   Or newline
        )p=g+y*n+c;                 //   Set current cell (happens first)
}