C ++在2D数组中找到两点之间的距离

Question

I'm working with inheritance right now, which I'm having no problem with. I have this vector of strings (basically a 2D array of char's).

The exact middle is where the shark is and the character of the shark changes based on where the other fish are.

For example

.....
.....
^.l..
.....
.....

The arrows are where the fish are and since the fish is to the left. The shark becomes a 'l'

.>...
.....
..u..
.....
.....

The arrows are where the fish are and since the fish is up. The shark becomes a 'u'

The code I had to do this correctly is this:

void Shark::point(std::vector<std::string>& map){

if (map[0][0] != '.' || map[1][0] != '.' || map[2][0] != '.' || map[3][0] != '.' || map[4][0] != '.' || map[1][1] != '.' || map[2][1] != '.' || map[3][1] != '.'){
ProtoFish::m_direction = Direction::left;
}
if (map[0][1] != '.' || map[0][2] != '.' || map[0][3] != '.' || map[1][2] != '.'){
ProtoFish::m_direction = Direction::up;
}
if (map[4][1] != '.' || map[4][2] != '.' || map[4][3] != '.' || map[3][2] != '.'){
ProtoFish::m_direction = Direction::down;
}
if (map[0][4] != '.' || map[1][4] != '.' || map[2][4] != '.' || map[3][4] != '.' || map[4][4] != '.' || map[1][3] != '.' || map[2][3] != '.' || map[3][3] != '.'){
ProtoFish::m_direction = Direction::right;
}
}

I am basically just checking if the fish are in this map:

    l u u u r
    l l u r r
    l l . r r
    l l d r r
    l d d d r

where if the '.' char doesn't exist that's where a fish is, so the Shark should point in that direction.

The problem occurs where there are multiple fish.

>....
..>..
..*..
.>...
..^>.

I understand what I'm supposed to do, but I have no idea of how to do it, or how to even get started. The goal is just to get the Shark facing the same direction of the nearest fish. So basically the arrow "closest" to the middle. Can someone help me out to get started with this? Maybe like a separate function that calculates the distance of each element from the center?

Answer 1

You need to calculate a distance for each fish - to do this you would convert your fish position to an (x,y) coordinate. Where the shark is is the (0,0) coordinate.

.....
.....
^.l..
.....
.....

would be (-2,0) for the fish ^

.>...
.....
..u..
.....
.....

would be (-1,3) for the fish >

so this coordinate conversion from 2D is easy to do. For the distance you use pythagoros theorem in 2 dimension-in this case its just square root of the square of x coordinate+square of y coordinate. You then need another new array to store the distances for each fish. Then you need to search that new distance array for smallest value where the index of the array is a fish index (like the id of fish). This will give you the nearest fish to the shark thus you can get get the shark facing the right way if there is one fist. However if there is more than one fish with the same distance, you need to think of a new rule in this case for example (could it be if there are more fish to left than right and the left and right fish are the same distance away-then face left beacause there is more fish?-or shark food-is a suggestion).