Hi so okay put it this way.
We have 12 players well what I want to do is find the closest one.
I've written some code below; two versions of my attempt (trust me theirs been a lot more, could of written a bible by now)
/*int client::closestEnemy()
{
for(u32 i = 0; i<12; i++)
{
DbgPrint("i: %i\n", i);
u32 index = 0;
if(!p[i].exist() || !p[i].alive() || p[i].team() == 3 || i == s.index)
continue;
for(u32 o = 0; o<12; o++)
if(vec.distance(*pos(), *p[i].pos()) <= vec.distance(*pos(), *p[o].pos()) || vec.distance(*pos(), *p[i].pos()) == vec.distance(*pos(), *p[o].pos()))
index++;
DbgPrint("PLAYER %s INDEX = %i\n", p[i].readName(), index);
if(index>10)
return i;
index = 0;
}
return -1;
}*/
int client::closestEnemy()
{
float distacnes[11] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
u32 index = 0;
for(u32 i = 0; i<12; i++)
if(p[i].exist() || p[i].alive() || p[i].team() != 3 || i != s.index)
distacnes[i] = vec.distance(*pos(), *p[i].pos());
else
continue;
for(u32 i = 0; i<12; i++)
if(distacnes[i]<distacnes[i+1])
index++;
DbgPrint("Score = %i\n", index);
if(index>11)
return index;
return -1;
}
What I am doing is simply increment index by 1 when ever another clients distance is larger then mine.
Now the first one is best bet but sometimes does not create a high enough index which makes zero sense because at least one client has to have a smaller distance then everyone else.
First, some indentation would help, I guess.
Then, there's no need to store all distances. You can keep track of the minimum distance encountered so far, like this:
int client::closestEnemy()
{
int index = -1;
float minDistance;
for(u32 i = 0; i<12; i++) {
if(p[i].exist() || p[i].alive() || p[i].team() != 3 || i != s.index) {
float distance = vec.distance(*pos(), *p[i].pos());
if (index == -1 || distance < minDistance) {
index = i;
minDistance = distance;
}
}
}
return index;
}
Related
The other day, I wrote a console game of Tic-Tac-Toe in c++ for my son. He wanted me to add a computer, and I ended us using the minimax algorithm for the first time. I did some quick testing, but really just gave my laptop to my son as soon as it was printing stuff, who played with it for a couple minuets. I looked over his sholder once or twice, and noticed that it wasn't playing optimally, iv'e been trying to debug it, but I can't see where it goes wrong. I tried getting rid of alpha beta prunning, but that did not change anything.
For context, on the board the computer is -1, blank is 0, and the player is 1.
Here is the minimax function:
int minimax(int board[9], int depth, int alpha, int beta, bool isMaxizimaizingPlayer)
{
bool found = false;
for (int i = 0; i < 9; i++)
{
if (board[i] == 0)
{
found = true;
}
}
if (!found)
{
return eval(board);
}
if (depth == 0 || eval(board) != 0)
{
return eval(board);
}
if (isMaxizimaizingPlayer)
{
int maxEval = -2;
for (int spot = 0; spot < 9; spot++)
{
if (board[spot] == 0)
{
board[spot] = 1;
int e = minimax(board, depth - 1, alpha, beta, false);
if (e > maxEval)
{
maxEval = e;
}
//if (beta < alpha)
//{
// break;
//}
board[spot] = 0;
}
}
return maxEval;
}
else {
int minEval = 2;
for (int spot = 0; spot < 9; spot++)
{
if (board[spot] == 0)
{
board[spot] = -1;
int e = minimax(board, depth - 1, alpha, beta, true);
if (e < minEval)
{
minEval = e;
}
//if (beta < alpha)
//{
// break;
//}
board[spot] = 0;
}
}
return minEval;
}
}
To be compleate, here is my eval function:
int eval(int board[9])
{
/*horizontial*/
for (int i = 0; i < 3; i++)
{
if (board[i * 3] == board[i * 3 + 1] && board[i * 3 + 2] == board[i * 3] && board[i * 3] != 0)
{
return board[i * 3];
}
}
/*vertical*/
for (int i = 0; i < 3; i++)
{
if (board[i] == board[i + 3] && board[i] == board[i + 6] && board[i] != 0)
{
return board[i];
}
}
/*Both diags*/
if (board[4] != 0) {
if (board[0] == board[4] && board[0] == board[8])
{
return board[4];
}
if (board[2] == board[4] && board[4] == board[6])
{
return board[4];
}
}
return 0;
}
And here is the inital call:
int spot = 0;
int minEval = 2;
for (int i = 0; i < 9; i++)
{
if (board[i] == 0)
{
board[i] = -1;
int score = minimax(board, 3, -2, 2, false);
if (score < minEval) {
minEval = score;
spot = i;
}
board[i] = 0;
}
}
std::cout << "The computer went in spot " << spot + 1 << std::endl;
board[spot] = -1;
printBoard(board);
It looks like you only call minimax with a depth of three, so the algorithm will only look up to three moves ahead, if you want optimal play you need to set the depth to > 9, so that the agent is always looking ahead to the end of the game.
I was solving Dungeon Game on LeetCode. While I was able to come up with the recurrence relation (and determine that it was a DP question in the first place), what is the intuition behind starting from the bottom right, as shown in a highly upvoted solution:
class Solution {
public:
int calculateMinimumHP(vector<vector<int> > &dungeon) {
int M = dungeon.size();
int N = dungeon[0].size();
// hp[i][j] represents the min hp needed at position (i, j)
// Add dummy row and column at bottom and right side
vector<vector<int> > hp(M + 1, vector<int>(N + 1, INT_MAX));
hp[M][N - 1] = 1;
hp[M - 1][N] = 1;
for (int i = M - 1; i >= 0; i--) {
for (int j = N - 1; j >= 0; j--) {
int need = min(hp[i + 1][j], hp[i][j + 1]) - dungeon[i][j];
hp[i][j] = need <= 0 ? 1 : need;
}
}
return hp[0][0];
}
};
I thought it to be pretty similar to the Unique Paths question, and hence started solving from the top left (but it doesn't retrieve correct results). My solution (if at all required) is:
class Solution {
public int calculateMinimumHP(int[][] dungeon) {
if(dungeon==null || dungeon.length==0 || dungeon[0]==null || dungeon[0].length==0)
return -1;
int maxVal = 0;
int[][] dp = new int[dungeon.length][dungeon[0].length];
dp[0][0] = dungeon[0][0];
for(int i=1; i<dungeon.length; i++)
dp[i][0] = dp[i-1][0] + dungeon[i][0];
for(int i=1; i<dungeon[0].length; i++)
dp[0][i] = dp[0][i-1] + dungeon[0][i];
int chosenOne=0;
for(int i=1; i<dungeon.length; i++) {
for(int j=1; j<dungeon[0].length; j++) {
if(dp[i-1][j] < 0 && dp[i][j-1] < 0) {
chosenOne = Math.max(dp[i-1][j], dp[i][j-1]);
}
else {
if(dp[i-1][j] > 0 && dp[i][j-1] > 0) {
chosenOne = Math.min(dp[i-1][j], dp[i][j-1]);
}
else {
chosenOne = (dp[i-1][j]>0?dp[i-1][j]:dp[i][j-1]);
}
}
dp[i][j] = dungeon[i][j] + chosenOne;
maxVal = Math.min(maxVal, dp[i][j]);
}
}
return maxVal+1;
}
}
One of the comments here by 'JaiMataDi' mentions that it is so because we don't know the initial health point of the knight. But how about the approach that I used?
To be precise, could some one please point out the intuition behind why we start at the bottom right and not the top left?
Thanks!
I'm trying to solve one problem, which I found on website https://open.kattis.com/problems/coast. Tl;dr version of problem is, that for given map of landscape, I should print out length of coastline (without inner islands).
My idea was, to solve this by adding additional layer and then start DFS, so the algorithm will walk through every possible tile in map, and then watch on every tile, how many borders are around the tile.
However, for specific input, is my algorithm not working. When I've submitted the solution on this site (open.kattis), it says, that my program is giving wrong answer in 9th of 26 tests (previous 8 test were ok), but without any further explanation.
Can somebody look at my program, and say me, why is it bad? Where did I do mistake? Thanks
#include <iostream>
#include <stack>
#include <sstream>
using namespace std;
int main() {
string line;
getline(cin, line);
int rows = 0;
int columns = 0;
stringstream stream(line);
stream >> rows;
stream >> columns;
int map[rows][columns];
for (int i = 0; i < rows; i++) {
getline(cin, line);
for (int j = 0; j < columns; j++) {
map[i][j] = line[j] - 48;
}
}
//parsed landscape into 2d array
// int rows = 5;
// int columns = 6;
// int map[rows][columns] = {
// {0, 1, 1, 1, 1, 0,},
// {0, 1, 0, 1, 1, 0,},
// {1, 1, 1, 0, 0, 0,},
// {0, 0, 0, 0, 1, 0,},
// {0, 0, 0, 0, 0, 0,},
// };
int bigMap[rows+2][columns+2];
bool visited[rows+2][columns+2];
//create bigger map, so DFS can start from corner and assume
//that there is water around everywhere
//also initialize array visited for DFS
//add 2 new rows, before and after existing one
for (int i = 0; i < columns+2; i++) {
bigMap[0][i] = 0;
bigMap[rows + 1][i] = 0;
visited[0][i] = false;
visited[rows + 1][i] = false;
}
//add 2 new columns, before and after existing
//copy original map to new one
for (int i = 0; i < rows; i++) {
bigMap[i+1][0] = 0;
bigMap[i+1][columns + 1] = 0;
visited[i+1][0] = false;
visited[i+1][columns + 1] = false;
for (int j = 0; j < columns; j++) {
bigMap[i+1][j+1] = map[i][j];
visited[i+1][j+1] = false;
}
}
rows += 2;
columns += 2;
//starting DFS
int x = 0, y = 0;
//visited[x][y] = true; <-- edit
pair <int, int> coordinates;
coordinates.first = x;
coordinates.second = y;
stack<pair <int, int> > st;
//first vertex in stack
st.push(coordinates);
//total sum of borders
int borders = 0;
while(!st.empty()) {
//check coordinates in each round
x = st.top().first;
y = st.top().second;
//navigate to new vertex (only if new vertex wasn't visited (visited[x][y] == 0) and only
//if there is water (bigMap[x][y] == 0) and check if new vertex is still in the map
//if there is no possible vertex, then we reached the end so then pop the vertex and
//look in another way
if (visited[x][y+1] == 0 && bigMap[x][y+1] == 0 && y + 1 < columns) {
y++;
coordinates.second = y;
st.push(coordinates);
} else {
if (visited[x+1][y] == 0 && bigMap[x+1][y] == 0 && x + 1 < rows) {
x++;
coordinates.first = x;
st.push(coordinates);
} else {
if (visited[x][y-1] == 0 && bigMap[x][y-1] == 0 && y > 0) {
y--;
coordinates.second = y;
st.push(coordinates);
} else {
if (visited[x-1][y] == 0 && bigMap[x-1][y] == 0 && x > 0) {
x--;
coordinates.first = x;
st.push(coordinates);
} else {
st.pop();
continue;
}
}
}
}
//visited new vertex, so look around him and count borders
visited[x][y] = true;
if (bigMap[x][y+1] == 1 && y + 1 < columns) borders++;
if (bigMap[x+1][y] == 1 && x + 1< rows) borders++;
if (bigMap[x][y-1] == 1 && y > 0) borders++;
if (bigMap[x-1][y] == 1 && x > 0) borders++;
}
cout << borders << endl;
return 0;
The issue is that you are reusing the variable coordinates each time around the loop without setting it to the correct value. Your if test cascade is assuming that coordinates is set to the current location. This is only true while you are descending in your dfs. Once you start ascending again, the coordinate will be pointing to the wrong place.
Simple solution, add
coordinates = st.top();
at the top of your loop.
Here is a sample map that it will currently get wrong.
5 6
000000
011100
001010
000100
000000
Answer should be 14, but currently you get 18 as the program reaches the lake at row 3, column 4.
To check that it is doing this, add a debugging line at the end of your loop, where it is adding the borders.
cout << "adding " << x << " " << y << "\n";
You can then verify if the program is considering any locations it shouldn't.
I think it will fail for {1,0,0,0},{0,1,1,0},{0,1,1,0},{0,0,0,0}. This is because the walk through is prevented from completing due to setting visited=true for vertex 0,0. Set it false for 0,0 instead should improve things. Hope it helps.
I was wondering how I can loop through a two dimentional array if the size of the array is random, e.g 6x6 or 10x10 etc. The idea is to search for four of the same kind of characters, 'x' or 'o'. This is typically needed for a board game.
int main() {
int array_size = 5; // Size of array
int array_height = array_size;
bool turn = true; // true = player 1, false = player 2
bool there_is_a_winner = false;
char** p_connect_four = new char*[array_size];
for (int i = 0; i < array_size; i++) // Initialise the 2D array
{ // At the same time set a value "_" as blank field
p_connect_four[i] = new char[array_size];
for (int j = 0; j < array_size; j++) {
p_connect_four[i][j] = '_';
}
}
}
This is what I have so far, checking from [3][0] to [0][3]. But this requires me to add 2 more for loops to check [4][0] to [0][4] and [4][1] to [1][4] IF the size of the board was 5x5.
for (int i = 3, j = 0; i > 0 && j < array_size; i--, j++ ) {// CHECK DOWN up right from 3,0 -> 0,3
if (p_connect_four[i][j] == p_connect_four[i - 1][j + 1] && p_connect_four[i][j] != '_' ) {
check_diagonalRight++;
if (check_diagonalRight == 3) {
there_is_a_winner = true;
break;
}
}
else {
check_diagonalRight = 0;
}
}
if (there_is_a_winner) { // Break while loop of game.
break;
}
Obviously I want to check the whole board diagonally to the right regardless of the size of the board. Is there any other way than having 3 separate for loops for checking
[3][0] -> [0][3] , [4][0] -> [0][4] and [4][1]-> [1][4] ?
for (i = array_size - 1, j = array_size - 2;
i < array_size && i >= 0, j < array_size && j >= 0; j--)
{ // starts from [4][3] and loops to the left if arraysize = 5x5
// but works on any size
int k = i, l = j;
for (k, l; k < array_size && k > 0, l < array_size && l > 0; k--, l++)
{ // checks diagonally to the right
if (check_diagonalRight == 3)
{
there_is_a_winner = true;
break;
}
if (p_connect_four[k][l] == p_connect_four[k - 1][l + 1] &&
p_connect_four[k][l] != '_')
{ //check up one square and right one square
check_diagonalRight++;
}
else
{
check_diagonalRight = 0;
// if its not equal, reset counter.
}
}
if (there_is_a_winner)
{
break; // break for loop
}
}
if (there_is_a_winner)
{
break; // break while loop of game
}
This checks up and right no matter the size, implement it for the other angles as well and it will work for any board size. You could potentially check right and left diagonal at once with nested loops.
This will work perfectly fine for your program! I hope so!
int arraySize = 8;
for(int i=0, j=0; i<arraySize && j<arraySize; i++, j++)
{
if((i == 0 && j == 0) || (i == arraySize - 1 && j == arraySize - 1))
{
continue;
}
else
{
int k = i;
int l = j;
//This Loop will check from central line (principal diagonal) to up right side (like slash sign / (representing direction))
for(k, l; k>0 && l < arraySize - 1; k--, l++)
{
//Here check your condition and increment to your variable. like:
if (p_connect_four[k][l] == p_connect_four[k - 1][l + 1] && p_connect_four[k][l] != '_' )
{
check_diagonalRight++;
}
}
//You can break the loop here if check_diagonalRight != k then break
k = i;
l = j;
//This Loop will check from central line (principal diagonal) to down left side (like slash sign / (representing direction))
for(k, l; k<arraySize - 1 && l > 0; k++, l--)
{
//Here check your condition and increment to your variable. like:
if (p_connect_four[k][l] == p_connect_four[k + 1][l - 1] && p_connect_four[k][l] != '_' )
{
check_diagonalRight++;
}
}
if(check_diagonalRight == i+j+1)
{
there_is_a_winner = true;
break;
}
}
}
I suggest to surround your board with extra special cases to avoid to check the bound.
To test each direction I suggest to use an array of offset to apply.
Following may help:
#include <vector>
using board_t = std::vector<std::vector<char>>;
constexpr const std::size_t MaxAlignment = 4;
enum Case {
Empty = '_',
X = 'X',
O = 'O',
Bound = '.'
};
enum class AlignmentResult { X, O, None };
// Create a new board, valid index would be [1; size] because of surrounding.
board_t new_board(std::size_t size)
{
// Create an empty board
board_t board(size + 2, std::vector<char>(size + 2, Case::Empty));
// Add special surround.
for (std::size_t i = 0; i != size + 2; ++i) {
board[0][i] = Case::Bound;
board[size + 1][i] = Case::Bound;
board[i][0] = Case::Bound;
board[i][size + 1] = Case::Bound;
}
return board_t;
}
// Test a winner from position in given direction.
AlignmentResult test(
const board_t& board,
std::size_t x, std::size_t y,
int offset_x, int offset_y)
{
if (board[x][y] == Case::Empty) {
return AlignmentResult::None;
}
for (std::size_t i = 1; i != MaxAlignment; ++i) {
// Following condition fails when going 'out of bound' thanks to Case::Bound,
// else you have also to check size...
if (board[x][y] != board[x + i * offset_x][y + i * offset_y]) {
return AlignmentResult::None;
}
}
if (board[x][y] == Case::X) {
return AlignmentResult::X;
} else {
return AlignmentResult::O;
}
}
// Test a winner on all the board
AlignmentResult test(const board_t& board)
{
// offset for direction. Use only 4 direction because of the symmetry.
const int offsets_x[] = {1, 1, 1, 0};
const int offsets_y[] = {-1, 0, 1, 1};
const std::size_t size = board.size() - 1;
for (std::size_t x = 1; x != size; ++x) {
for (std::size_t y = 1; y != size; ++y) {
for (std::size_t dir = 0; dir != 4; ++dir) { // for each directions
auto res = test(board, x, y, offsets_x[dir], offsets_y[y]);
if (res != AlignmentResult::None) {
return res;
}
}
}
}
return AlignmentResult::None;
}
I need to place numbers within a grid such that it doesn't collide with each other. This number placement should be random and can be horizontal or vertical. The numbers basically indicate the locations of the ships. So the points for the ships should be together and need to be random and should not collide.
I have tried it:
int main()
{
srand(time(NULL));
int Grid[64];
int battleShips;
bool battleShipFilled;
for(int i = 0; i < 64; i++)
Grid[i]=0;
for(int i = 1; i <= 5; i++)
{
battleShips = 1;
while(battleShips != 5)
{
int horizontal = rand()%2;
if(horizontal == 0)
{
battleShipFilled = false;
while(!battleShipFilled)
{
int row = rand()%8;
int column = rand()%8;
while(Grid[(row)*8+(column)] == 1)
{
row = rand()%8;
column = rand()%8;
}
int j = 0;
if(i == 1) j= (i+1);
else j= i;
for(int k = -j/2; k <= j/2; k++)
{
int numberOfCorrectLocation = 0;
while(numberOfCorrectLocation != j)
{
if(row+k> 0 && row+k<8)
{
if(Grid[(row+k)*8+(column)] == 1) break;
numberOfCorrectLocation++;
}
}
if(numberOfCorrectLocation !=i) break;
}
for(int k = -j/2; k <= j/2; k++)
Grid[(row+k)*8+(column)] = 1;
battleShipFilled = true;
}
battleShips++;
}
else
{
battleShipFilled = false;
while(!battleShipFilled)
{
int row = rand()%8;
int column = rand()%8;
while(Grid[(row)*8+(column)] == 1)
{
row = rand()%8;
column = rand()%8;
}
int j = 0;
if(i == 1) j= (i+1);
else j= i;
for(int k = -j/2; k <= j/2; k++)
{
int numberOfCorrectLocation = 0;
while(numberOfCorrectLocation != i)
{
if(row+k> 0 && row+k<8)
{
if(Grid[(row)*8+(column+k)] == 1) break;
numberOfCorrectLocation++;
}
}
if(numberOfCorrectLocation !=i) break;
}
for(int k = -j/2; k <= j/2; k++)
Grid[(row)*8+(column+k)] = 1;
battleShipFilled = true;
}
battleShips++;
}
}
}
}
But the code i have written is not able to generate the numbers randomly in the 8x8 grid.
Need some guidance on how to solve this. If there is any better way of doing it, please tell me...
How it should look:
What My code is doing:
Basically, I am placing 5 ships, each of different size on a grid. For each, I check whether I want to place it horizontally or vertically randomly. After that, I check whether the surrounding is filled up or not. If not, I place them there. Or I repeat the process.
Important Point: I need to use just while, for loops..
You are much better of using recursion for that problem. This will give your algorithm unwind possibility. What I mean is that you can deploy each ship and place next part at random end of the ship, then check the new placed ship part has adjacent tiles empty and progress to the next one. if it happens that its touches another ship it will due to recursive nature it will remove the placed tile and try on the other end. If the position of the ship is not valid it should place the ship in different place and start over.
I have used this solution in a word search game, where the board had to be populated with words to look for. Worked perfect.
This is a code from my word search game:
bool generate ( std::string word, BuzzLevel &level, CCPoint position, std::vector<CCPoint> &placed, CCSize lSize )
{
std::string cPiece;
if ( word.size() == 0 ) return true;
if ( !level.inBounds ( position ) ) return false;
cPiece += level.getPiece(position)->getLetter();
int l = cPiece.size();
if ( (cPiece != " ") && (word[0] != cPiece[0]) ) return false;
if ( pointInVec (position, placed) ) return false;
if ( position.x >= lSize.width || position.y >= lSize.height || position.x < 0 || position.y < 0 ) return false;
placed.push_back(position);
bool used[6];
for ( int t = 0; t < 6; t++ ) used[t] = false;
int adj;
while ( (adj = HexCoord::getRandomAdjacentUnique(used)) != -1 )
{
CCPoint nextPosition = HexCoord::getAdjacentGridPositionInDirection((eDirection) adj, position);
if ( generate ( word.substr(1, word.size()), level, nextPosition, placed, lSize ) ) return true;
}
placed.pop_back();
return false;
}
CCPoint getRandPoint ( CCSize size )
{
return CCPoint ( rand() % (int)size.width, rand() % (int)size.height);
}
void generateWholeLevel ( BuzzLevel &level,
blockInfo* info,
const CCSize &levelSize,
vector<CCLabelBMFont*> wordList
)
{
for ( vector<CCLabelBMFont*>::iterator iter = wordList.begin();
iter != wordList.end(); iter++ )
{
std::string cWord = (*iter)->getString();
// CCLog("Curront word %s", cWord.c_str() );
vector<CCPoint> wordPositions;
int iterations = 0;
while ( true )
{
iterations++;
//CCLog("iteration %i", iterations );
CCPoint cPoint = getRandPoint(levelSize);
if ( generate (cWord, level, cPoint, wordPositions, levelSize ) )
{
//Place pieces here
for ( int t = 0; t < cWord.size(); t++ )
{
level.getPiece(wordPositions[t])->addLetter(cWord[t]);
}
break;
}
if ( iterations > 1500 )
{
level.clear();
generateWholeLevel(level, info, levelSize, wordList);
return;
}
}
}
}
I might add that shaped used in the game was a honeycomb. Letter could wind in any direction, so the code above is way more complex then what you are looking for I guess, but will provide a starting point.
I will provide something more suitable when I get back home as I don't have enough time now.
I can see a potential infinite loop in your code
int j = 0;
if(i == 1) j= (i+1);
else j= i;
for(int k = -j/2; k <= j/2; k++)
{
int numberOfCorrectLocation = 0;
while(numberOfCorrectLocation != i)
{
if(row+k> 0 && row+k<8)
{
if(Grid[(row)*8+(column+k)] == 1) break;
numberOfCorrectLocation++;
}
}
if(numberOfCorrectLocation !=i) break;
}
Here, nothing prevents row from being 0, as it was assignd rand%8 earlier, and k can be assigned a negative value (since j can be positive). Once that happens nothing will end the while loop.
Also, I would recommend re-approaching this problem in a more object oriented way (or at the very least breaking up the code in main() into multiple, shorter functions). Personally I found the code a little difficult to follow.
A very quick and probably buggy example of how you could really clean your solution up and make it more flexible by using some OOP:
enum Orientation {
Horizontal,
Vertical
};
struct Ship {
Ship(unsigned l = 1, bool o = Horizontal) : length(l), orientation(o) {}
unsigned char length;
bool orientation;
};
class Grid {
public:
Grid(const unsigned w = 8, const unsigned h = 8) : _w(w), _h(h) {
grid.resize(w * h);
foreach (Ship * sp, grid) {
sp = nullptr;
}
}
bool addShip(Ship * s, unsigned x, unsigned y) {
if ((x <= _w) && (y <= _h)) { // if in valid range
if (s->orientation == Horizontal) {
if ((x + s->length) <= _w) { // if not too big
int p = 0; //check if occupied
for (int c1 = 0; c1 < s->length; ++c1) if (grid[y * _w + x + p++]) return false;
p = 0; // occupy if not
for (int c1 = 0; c1 < s->length; ++c1) grid[y * _w + x + p++] = s;
return true;
} else return false;
} else {
if ((y + s->length) <= _h) {
int p = 0; // check
for (int c1 = 0; c1 < s->length; ++c1) {
if (grid[y * _w + x + p]) return false;
p += _w;
}
p = 0; // occupy
for (int c1 = 0; c1 < s->length; ++c1) {
grid[y * _w + x + p] = s;
p += _w;
}
return true;
} else return false;
}
} else return false;
}
void drawGrid() {
for (int y = 0; y < _h; ++y) {
for (int x = 0; x < _w; ++x) {
if (grid.at(y * w + x)) cout << "|S";
else cout << "|_";
}
cout << "|" << endl;
}
cout << endl;
}
void hitXY(unsigned x, unsigned y) {
if ((x <= _w) && (y <= _h)) {
if (grid[y * _w + x]) cout << "You sunk my battleship" << endl;
else cout << "Nothing..." << endl;
}
}
private:
QVector<Ship *> grid;
unsigned _w, _h;
};
The basic idea is create a grid of arbitrary size and give it the ability to "load" ships of arbitrary length at arbitrary coordinates. You need to check if the size is not too much and if the tiles aren't already occupied, that's pretty much it, the other thing is orientation - if horizontal then increment is +1, if vertical increment is + width.
This gives flexibility to use the methods to quickly populate the grid with random data:
int main() {
Grid g(20, 20);
g.drawGrid();
unsigned shipCount = 20;
while (shipCount) {
Ship * s = new Ship(qrand() % 8 + 2, qrand() %2);
if (g.addShip(s, qrand() % 20, qrand() % 20)) --shipCount;
else delete s;
}
cout << endl;
g.drawGrid();
for (int i = 0; i < 20; ++i) g.hitXY(qrand() % 20, qrand() % 20);
}
Naturally, you can extend it further, make hit ships sink and disappear from the grid, make it possible to move ships around and flip their orientation. You can even use diagonal orientation. A lot of flexibility and potential to harness by refining an OOP based solution.
Obviously, you will put some limits in production code, as currently you can create grids of 0x0 and ships of length 0. It's just a quick example anyway. I am using Qt and therefore Qt containers, but its just the same with std containers.
I tried to rewrite your program in Java, it works as required. Feel free to ask anything that is not clearly coded. I didn't rechecked it so it may have errors of its own. It can be further optimized and cleaned but as it is past midnight around here, I would rather not do that at the moment :)
public static void main(String[] args) {
Random generator = new Random();
int Grid[][] = new int[8][8];
for (int battleShips = 0; battleShips < 5; battleShips++) {
boolean isHorizontal = generator.nextInt(2) == 0 ? true : false;
boolean battleShipFilled = false;
while (!battleShipFilled) {
// Select a random row and column for trial
int row = generator.nextInt(8);
int column = generator.nextInt(8);
while (Grid[row][column] == 1) {
row = generator.nextInt(8);
column = generator.nextInt(8);
}
int lengthOfBattleship = 0;
if (battleShips == 0) // Smallest ship should be of length 2
lengthOfBattleship = (battleShips + 2);
else // Other 4 ships has the length of 2, 3, 4 & 5
lengthOfBattleship = battleShips + 1;
int numberOfCorrectLocation = 0;
for (int k = 0; k < lengthOfBattleship; k++) {
if (isHorizontal && row + k > 0 && row + k < 8) {
if (Grid[row + k][column] == 1)
break;
} else if (!isHorizontal && column + k > 0 && column + k < 8) {
if (Grid[row][column + k] == 1)
break;
} else {
break;
}
numberOfCorrectLocation++;
}
if (numberOfCorrectLocation == lengthOfBattleship) {
for (int k = 0; k < lengthOfBattleship; k++) {
if (isHorizontal)
Grid[row + k][column] = 1;
else
Grid[row][column + k] = 1;
}
battleShipFilled = true;
}
}
}
}
Some important points.
As #Kindread said in an another answer, the code has an infinite loop condition which must be eliminated.
This algorithm will use too much resources to find a solution, it should be optimized.
Code duplications should be avoided as it will result in more maintenance cost (which might not be a problem for this specific case), and possible bugs.
Hope this answer helps...