Java 拼图游戏Android-DFS算法
我有一个叫做Islands and bridges的android应用程序,也称为 该应用程序使用一个二维数组,每当用户重新启动游戏时,该数组都会随机生成岛屿。它形成一个数字为0到4的矩阵,其中0=null,1-4=岛屿。可以有两座桥从一个岛屿连接到另一个岛屿,目前地图无法解。为了解决游戏,用户需要使用网桥连接岛屿,因此,如果岛屿=4,则需要4个连接;如果岛屿=2,则需要2个连接,依此类推 在我的研究中,我发现解决游戏的最佳算法是使用深度优先搜索- 我已经研究了上的不同问题,但似乎找不到解决方案,因为我的数组是Java 拼图游戏Android-DFS算法,java,android,algorithm,depth-first-search,Java,Android,Algorithm,Depth First Search,我有一个叫做Islands and bridges的android应用程序,也称为 该应用程序使用一个二维数组,每当用户重新启动游戏时,该数组都会随机生成岛屿。它形成一个数字为0到4的矩阵,其中0=null,1-4=岛屿。可以有两座桥从一个岛屿连接到另一个岛屿,目前地图无法解。为了解决游戏,用户需要使用网桥连接岛屿,因此,如果岛屿=4,则需要4个连接;如果岛屿=2,则需要2个连接,依此类推 在我的研究中,我发现解决游戏的最佳算法是使用深度优先搜索- 我已经研究了上的不同问题,但似乎找不到解决方案
String
类型,而不是integer
问题如何应用DFS算法连接岛屿
这是我的应用程序的屏幕截图
此函数用于创建easy map 4x4矩阵:
private void InitializeEasy() {
Random rand = new Random();
String[][] debug_board_state = new String[4][4];
setCurrentState(new State(WIDTH_EASY));
for (int row = 0; row < debug_board_state.length; row++) {
for (int column = 0; column < debug_board_state[row].length; column++) {
debug_board_state[row][column] = String.valueOf(rand.nextInt(5));
}
}
for (int row = 0; row < debug_board_state.length; row++) {
for (int column = 0; column < debug_board_state[row].length; column++) {
System.out.print(debug_board_state[row][column] + " ");
}
System.out.println();
}
for (int row = 0; row < WIDTH_EASY; ++row) {
for (int column = 0; column < WIDTH_EASY; ++column) {
for (int colNum = column - 1; colNum <= (column + 1); colNum += 1) {
getCurrentState().board_elements[row][column] = new BoardElement();
getCurrentState().board_elements[row][column].max_connecting_bridges = Integer.parseInt(debug_board_state[row][column]);
getCurrentState().board_elements[row][column].row = row;
getCurrentState().board_elements[row][column].col = column;
if (getCurrentState().board_elements[row][column].max_connecting_bridges > 0) {
getCurrentState().board_elements[row][column].is_island = true;
}
}
}
}
}
private void InitializeEasy(){
Random rand=新的Random();
字符串[][]调试板状态=新字符串[4][4];
setCurrentState(新状态(宽度);
for(int row=0;row
DFS可应用于游戏状态
伪算法:
int[][]起始\u线索={
{2, 0, 0, 3, 0, 3},
{0, 1, 4, 0, 4, 0},
{0, 0, 0, 0, 0, 0},
{3, 0, 3, 0, 2, 0},
{0, 0, 0, 1, 0, 2},
{2, 0, 4, 0, 2, 0}
};
无效搜索(){
Map remainingOptions=new HashMap();
Stack gameTree=新堆栈();
gameTree.push(新的土地(开始的线索));
while(true){
landstate=gameTree.peek();
int[]p=state.lowestTodo();
if(p==null)
System.out.println(“找到解决方案”);
//移动到下一个游戏状态
int r=p[0];
int c=p[1];
System.out.println(“在(“+r+”,“+c+”)处扩展节点的游戏状态”);
列表ds=null;
if(剩余选项containsKey(新点(r,c)))
ds=剩余的选项.get(新点(r,c));
否则{
ds=新的ArrayList();
for(Direction dir:Direction.values()){
int[]tmp=state.nextIsland(r,c,dir);
if(tmp==null)
继续;
if(state.canBuildBridge(r,c,tmp[0],tmp[1]))
ds.add(dir);;
}
保留选项。放置(新点(r,c),ds);
}
//如果节点无法再扩展,并且无法回溯,我们将退出
if(ds.isEmpty()&&gameTree.isEmpty()){
System.out.println(“未找到有效配置”);
返回;
}
//如果节点不能再扩展,我们需要回溯
if(ds.isEmpty()){
gameTree.pop();
删除(新点(r,c));
System.out.println(“回到以前的决定”);
继续;
}
方向dir=ds.remove(0);
System.out.println(“连接”+dir.name());
保留选项。放置(新点(r,c),ds);
土地下一州=新土地(州);
int[]tmp=state.nextIsland(r,c,dir);
nextState.connect(r,c,tmp[0],tmp[1]);
gameTree.push(nextState);
}
}
公共静态void main(字符串[]args){
新建Main().search();
}
我还编写了一个实用程序类,用于处理需要在其上建造桥梁的这块土地上的常见操作(如查找下一个可用的岛屿、检查是否可以建造桥梁等)
公共类土地{
私人int[][]为您建造桥梁;
私有布尔值[][]是_岛;
私人方向[]桥梁已建成;
公共土地(国际[][]布里奇斯托多){
BRIDGES_TO_BUILD=复制(bridgesToDo);
int R=bridgesToDo.length;
int C=bridgesToDo[0]。长度;
已建桥梁=新方向[R][C];
IS_ISLAND=新布尔值[R][C];
对于(int i=0;i 0;
}
}
}
公共土地(其他土地){
BRIDGES_TO_BUILD=复制(其他.BRIDGES_TO_BUILD);
int R=桥梁至建筑长度;
int C=桥梁到建筑[0]。长度;
已建桥梁=新方向[R][C];
IS_ISLAND=新布尔值[R][C];
对于(int i=0;i=R | | c<0 | | c>=c)
返回null;
//运动矢量
int[][]运动向量={{-1,0},{0,1},{1,0},{0,-1};
int i=Arrays.asList(Direction.value
int[][] STARTING_CLUES = {
{2, 0, 0, 3, 0, 3},
{0, 1, 4, 0, 4, 0},
{0, 0, 0, 0, 0, 0},
{3, 0, 3, 0, 2, 0},
{0, 0, 0, 1, 0, 2},
{2, 0, 4, 0, 2, 0}
};
void search(){
Map<Point, List<Direction>> remainingOptions = new HashMap<>();
Stack<Land> gameTree = new Stack<>();
gameTree.push(new Land(STARTING_CLUES));
while(true){
Land state = gameTree.peek();
int[] p = state.lowestTodo();
if (p == null)
System.out.println("solution found");
// move to next game state
int r = p[0];
int c = p[1];
System.out.println("expanding game state for node at (" + r + ", " + c + ")");
List<Direction> ds = null;
if(remainingOptions.containsKey(new Point(r,c)))
ds = remainingOptions.get(new Point(r,c));
else{
ds = new ArrayList<>();
for(Direction dir : Direction.values()) {
int[] tmp = state.nextIsland(r, c, dir);
if(tmp == null)
continue;
if(state.canBuildBridge(r,c,tmp[0], tmp[1]))
ds.add(dir);
}
remainingOptions.put(new Point(r,c), ds);
}
// if the node can no longer be expanded, and backtracking is not possible we quit
if(ds.isEmpty() && gameTree.isEmpty()){
System.out.println("no valid configuration found");
return;
}
// if the node can no longer be expanded, we need to backtrack
if(ds.isEmpty()){
gameTree.pop();
remainingOptions.remove(new Point(r,c));
System.out.println("going back to previous decision");
continue;
}
Direction dir = ds.remove(0);
System.out.println("connecting " + dir.name());
remainingOptions.put(new Point(r,c), ds);
Land nextState = new Land(state);
int[] tmp = state.nextIsland(r,c,dir);
nextState.connect(r,c, tmp[0], tmp[1]);
gameTree.push(nextState);
}
}
public static void main(String[] args) {
new Main().search();
}
public class Land {
private int[][] BRIDGES_TO_BUILD;
private boolean[][] IS_ISLAND;
private Direction[][] BRIDGES_ALREADY_BUILT;
public Land(int[][] bridgesToDo){
BRIDGES_TO_BUILD = copy(bridgesToDo);
int R = bridgesToDo.length;
int C = bridgesToDo[0].length;
BRIDGES_ALREADY_BUILT = new Direction[R][C];
IS_ISLAND = new boolean[R][C];
for(int i=0;i<R;i++) {
for (int j = 0; j < C; j++) {
BRIDGES_ALREADY_BUILT[i][j] = null;
IS_ISLAND[i][j] = bridgesToDo[i][j] > 0;
}
}
}
public Land(Land other){
BRIDGES_TO_BUILD = copy(other.BRIDGES_TO_BUILD);
int R = BRIDGES_TO_BUILD.length;
int C = BRIDGES_TO_BUILD[0].length;
BRIDGES_ALREADY_BUILT = new Direction[R][C];
IS_ISLAND = new boolean[R][C];
for(int i=0;i<R;i++) {
for (int j = 0; j < C; j++) {
BRIDGES_ALREADY_BUILT[i][j] = other.BRIDGES_ALREADY_BUILT[i][j];
IS_ISLAND[i][j] = other.IS_ISLAND[i][j];
}
}
}
public int[] next(int r, int c, Direction dir){
int R = BRIDGES_TO_BUILD.length;
int C = BRIDGES_TO_BUILD[0].length;
// out of bounds
if(r < 0 || r >=R || c < 0 || c >= C)
return null;
// motion vectors
int[][] motionVector = {{-1, 0},{0,1},{1,0},{0,-1}};
int i = Arrays.asList(Direction.values()).indexOf(dir);
// calculate next
int[] out = new int[]{r + motionVector[i][0], c + motionVector[i][1]};
r = out[0];
c = out[1];
// out of bounds
if(r < 0 || r >=R || c < 0 || c >= C)
return null;
// return
return out;
}
public int[] nextIsland(int r, int c, Direction dir){
int[] tmp = next(r,c,dir);
if(tmp == null)
return null;
while(!IS_ISLAND[tmp[0]][tmp[1]]){
tmp = next(tmp[0], tmp[1], dir);
if(tmp == null)
return null;
}
return tmp;
}
public boolean canBuildBridge(int r0, int c0, int r1, int c1){
if(r0 == r1 && c0 > c1){
return canBuildBridge(r0, c1, r1, c0);
}
if(c0 == c1 && r0 > r1){
return canBuildBridge(r1, c0, r0, c1);
}
if(r0 == r1){
int[] tmp = nextIsland(r0, c0, Direction.EAST);
if(tmp[0] != r1 || tmp[1] != c1)
return false;
if(BRIDGES_TO_BUILD[r0][c0] == 0)
return false;
if(BRIDGES_TO_BUILD[r1][c1] == 0)
return false;
for (int i = c0; i <= c1 ; i++) {
if(IS_ISLAND[r0][i])
continue;
if(BRIDGES_ALREADY_BUILT[r0][i] == Direction.NORTH)
return false;
}
}
if(c0 == c1){
int[] tmp = nextIsland(r0, c0, Direction.SOUTH);
if(tmp[0] != r1 || tmp[1] != c1)
return false;
if(BRIDGES_TO_BUILD[r0][c0] == 0 || BRIDGES_TO_BUILD[r1][c1] == 0)
return false;
for (int i = r0; i <= r1 ; i++) {
if(IS_ISLAND[i][c0])
continue;
if(BRIDGES_ALREADY_BUILT[i][c0] == Direction.EAST)
return false;
}
}
// default
return true;
}
public int[] lowestTodo(){
int R = BRIDGES_TO_BUILD.length;
int C = BRIDGES_TO_BUILD[0].length;
int[] out = {0, 0};
for (int i=0;i<R;i++) {
for (int j = 0; j < C; j++) {
if(BRIDGES_TO_BUILD[i][j] == 0)
continue;
if (BRIDGES_TO_BUILD[out[0]][out[1]] == 0)
out = new int[]{i, j};
if (BRIDGES_TO_BUILD[i][j] < BRIDGES_TO_BUILD[out[0]][out[1]])
out = new int[]{i, j};
}
}
if (BRIDGES_TO_BUILD[out[0]][out[1]] == 0) {
return null;
}
return out;
}
private int[][] copy(int[][] other){
int[][] out = new int[other.length][other.length == 0 ? 0 : other[0].length];
for(int r=0;r<other.length;r++)
out[r] = Arrays.copyOf(other[r], other[r].length);
return out;
}
public void connect(int r0, int c0, int r1, int c1){
if(r0 == r1 && c0 > c1){
connect(r0, c1, r1, c0);
return;
}
if(c0 == c1 && r0 > r1){
connect(r1, c0, r0, c1);
return;
}
if(!canBuildBridge(r0, c0, r1, c1))
return;
BRIDGES_TO_BUILD[r0][c0]--;
BRIDGES_TO_BUILD[r1][c1]--;
if(r0 == r1){
for (int i = c0; i <= c1 ; i++) {
if(IS_ISLAND[r0][i])
continue;
BRIDGES_ALREADY_BUILT[r0][i] = Direction.EAST;
}
}
if(c0 == c1){
for (int i = r0; i <= r1 ; i++) {
if(IS_ISLAND[i][c0])
continue;
BRIDGES_ALREADY_BUILT[i][c0] = Direction.NORTH;
}
}
}
}