我正在努力制定一个合适的算法，以便从迷宫中的 START 位置转到 EXIT 位置。 值得一提的是，迷宫是矩形， maxsize 500x500 ，理论上，DFS可以通过一些分支和绑定技术来解析......
10 3 4
7 6
3 3 1 2 2 1 0
2 2 2 4 2 2 5
2 2 1 3 0 2 2
2 2 1 3 3 4 2
3 4 4 3 1 1 3
1 2 2 4 2 2 1
Output:
5 1 4 2
说明：
我们的经纪人每走一步就会失去能量，他只能向上，向下，向左和向右移动。此外，如果代理人的剩余能量为零或更少，他就会死亡，所以我们打印出类似“不可能”的东西。
因此，在输入 10 是初始代理的能量， 3 4 是 START 位置（即第3列，第4行）我们有一个迷宫 7x6 。把它想象成一种迷宫，在这种迷宫中，我想找到一个让代理人有更好的剩余能量（最短路径）的出口。
如果存在导致相同剩余能量的路径，我们选择步数较少的路径。
我需要知道在最坏的情况下DFS到迷宫500x500是否可行，这些限制以及如何操作，存储每个步骤中的剩余能量以及到目前为止所采取的步骤数。
输出意味着代理以剩余能量= 5到达出口位置4以2个步骤到达。如果我们仔细观察，在这个迷宫中，它也可以以相同的能量但以3个步骤退出位置3 1（第3列，第1行），因此我们选择更好的一个。
考虑到这些，有人可以帮我一些代码或伪代码吗？ 我在使用2D阵列时遇到了麻烦，以及如何存储剩余的能量，路径（或采取的步骤数）....
编辑：
拉里，就像我说的那样，我对这段代码感到有点困惑。这是我到目前为止所尝试的，只是为了确定从START到EXIT步骤较少的最短路径，同时修复EXIT ......public class exitFromMaze {
int energy, startY, startX, xMax, yMax;
int adjMatrix[][];
boolean visited[][];
ArrayList<Cell> neighbours;
//ArrayList<Cell> visited;
Cell start;
Stack<Cell> stack;
public exM() {
Scanner cin = new Scanner(System.in);
int nrTests = cin.nextInt();
for (int i = 0; i < nrTests; i++) {
energy = cin.nextInt();
startY = cin.nextInt()-1; //start at columnstartY
startX = cin.nextInt()-1; //start at line startX
xMax = cin.nextInt();//7 cols
yMax = cin.nextInt(); //8 rows
adjMatrix = new int[yMax][xMax];
visited = new boolean[yMax][xMax];
//visited = new ArrayList<Cell>();
this.stack = new Stack<Cell>();
for (int r = 0; r < yMax; r++) { // yMax linhas
for (int c = 0; c < xMax; c++) { // xMax colunas
adjMatrix[r][c] = cin.nextInt();
visited[r][c] = false;
//System.out.println("matrix["+r+"]["+c+"] = "+adjMatrix[r][c]);
}
}
start= new Cell(startX, startY, 0);
//adiciona a pos actual à pilha de celulas/nos
stack.push(start);
//printArray(yMax, xMax);
findBestExit();
}//end_of_test_Cases
}
private void findBestExit() {
// BEGINNING OF DEPTH-FIRST SEARCH
Cell curCell;
while (!(stack.empty())) {
curCell = (Cell) (stack.pop());
//just fix an exit point ...for now (if it works for one, it has to work for all the other possible exits)
if (curCell.row==0 && curCell.col== 4) {
System.out.println("Arrived at pos: "+curCell.row+","+curCell.col+" with E= "+(energy-curCell.getEnergy())+" with "+curCell.getSteps()+" steps");
//finish = curCell;
break;
} else {
visited[curCell.row][curCell.col] = true;
}
this.neighbours = (ArrayList<Cell>) curCell.getNeighbours(this.xMax, this.yMax);
for (Cell neighbourCell: neighbours) {
//1- I think something's missing here and it would be here the point to cut some cases...isn't it?
if ( curCell.getEnergy() + neighbourCell.getEnergy() < this.energy && !visited[neighbourCell.row][neighbourCell.col]){
neighbourCell.energy+= curCell.energy;
neighbourCell.setSteps(curCell.getSteps()+1);
neighbourCell.setPrevious(curCell);
stack.push(neighbourCell);
}
// ...
}
}
// END OF DEPTH-FIRST SEARCH and DIJKSTRA?
}
class Cell {
int row;
int col;
int energy;
int steps;
Cell previous;
//Node next;
public Cell(int x, int y, int steps) {
this.row = x;
this.col = y;
this.energy = adjMatrix[x][y];
this.steps = steps;
//this.next = null;
this.previous = null;
}
public Cell(int x, int y, Cell prev) {
this.row = x;
this.col = y;
this.steps = 0;
this.energy = adjMatrix[x][y];
this.previous = prev;
}
@Override
public String toString() {
return "(,"+this.getRow()+","+this.getCol()+")";
}
public int getEnergy() {
return energy;
}
public void setEnergy(int energy) {
this.energy = energy;
}
public Cell getPrevious() {
return previous;
}
public void setPrevious(Cell previous) {
this.previous = previous;
}
public int getRow() {
return row;
}
public void setRow(int x) {
this.row = x;
}
public int getCol() {
return col;
}
public void setCol(int y) {
this.col = y;
}
public int getSteps() {
return steps;
}
public void setSteps(int steps) {
this.steps = steps;
}
public Cell south(int verticalLimit) {
Cell ret = null;
if (row < (verticalLimit - 1)) {
ret = new Cell(row+1, col, this);
//ret.previous = this;
}
return ret;
}
/**
* Gives the north to our current Cell
* @return the Cell in that direction, null if it's impossible
* to go in that direction
*/
public Cell north() {
Cell ret = null;
if (row > 0) {
ret = new Cell(row-1, col ,this);
//ret.previous = this;
}
return ret;
}
/**
* Gives the west (left) to our current Cell
* @return the Cell in that direction, null if it's
* impossible to go in that direction
*/
public Cell west() {
Cell ret = null;
if (col > 0) {
ret = new Cell(row, col-1,this);
//ret.previous = this;
}
return ret;
}
/**
* Gives the east direction(right) to our current Cell
* @return the Cell in that direction, null if it's
* impossible to go in that direction
*/
public Cell east(int horizontalLimit) {
Cell ret = null;
//if it's inside the number max of collumns
if (col < (horizontalLimit - 1)) {
ret = new Cell(row , col+1, this);
}
return ret;
}
public List getNeighbours(int xlimit, int ylimit) {
ArrayList<Cell> res = new ArrayList<Cell>(4);
Cell n;
n = south(ylimit);
if (n != null) {
res.add(n);
}
n = north();
if (n != null) {
res.add(n);
}
n = east(xlimit);
if (n != null) {
res.add(n);
}
n = west();
if (n != null) {
res.add(n);
}
return res;
}
}
private void printArray(int h, int w) {
int i, j;
// print array in rectangular form
System.out.print(" ");
for (i = 0; i < w; i++) {
System.out.print("\t" + i);
}
System.out.println();
for (int r = 0; r < h; r++) {
System.out.print(" " + r);
for (int c = 0; c < w; c++) {
System.out.print("\t" + adjMatrix[r][c]);
}
System.out.println("");
}
System.out.println();
}
public static void main(String args[]) {
new exM();
}
}
输入：
1
40 3 3
7 8
12 11 12 11 3 12 12
12 11 11 12 2 1 13
11 11 12 2 13 2 14
10 11 13 3 2 1 12
10 11 13 13 11 12 13
12 12 11 13 11 13 12
13 12 12 11 11 11 11
13 13 10 10 13 11 12
应该打印：
12 5 1 8
，即代理退出更好的退出，（0,4），剩余能量= 12，只有8步。
根据我的想法，你的帮助，是否要求我指出失败或纠正错误？ 我对此感到厌倦......因为我必须让事情变得复杂......
更多输入/输出（当无法实现退出时（使用Energy＆gt; 0），只需打印该事实）。
3
40 3 3
7 8
12 11 12 11 3 12 12
12 11 11 12 2 1 13
11 11 12 2 13 2 14
10 11 13 3 2 1 12
10 11 13 13 11 12 13
12 12 11 13 11 13 12
13 12 12 11 11 11 11
13 13 10 10 13 11 12
8 3 4
7 6
4 3 3 2 2 3 2
2 5 2 2 2 3 3
2 1 2 2 3 2 2
4 3 3 2 2 4 1
3 1 4 3 2 3 1
2 2 3 3 0 3 4
10 3 4
7 6
3 3 1 2 2 1 0
2 2 2 4 2 2 5
2 2 1 3 0 2 2
2 2 1 3 3 4 2
3 4 4 3 1 1 3
1 2 2 4 2 2 1
Output
12 5 1 8
Goodbye cruel world!
5 1 4 2
答案 0 :(得分：8)
使用Dijkstra's algorithm，使用基本方向中的隐式图表。使用堆实现它将是O(V log V)，这应该足够500x500。第一次放松节点时，它使用的能量最低，你可以到达那里。您可以使用此算法相当简单地设置节点的前驱。
编辑：一些伪代码，解释了Dijkstra的算法：
function Dijkstra( graph, source ):
// distance is infinity to everywhere initially
dist = initialize list of size V to infinity
// no vertices pointed to anything
previous = initialize list of size V to undefined
// distance from source to source is 0
dist[source] = 0
Q = priority queue
insert all vertices into Q
while Q is not empty:
// get the vertex closest to the source
current_vertex = Q.pop
if dist[current_vertex] == infinity
break
// these are the adjacent vertices in the four cardinal direction
for each vertex next to current_vertex:
// if it costs less energy to go to vertex
// from current_vertex
if ( dist[current_vertex] + energy[vertex] < dist[vertex] )
dist[vertex] = dist[current_vertex] + energy[vertex]
previous[vertex] = current_vertex
// Another if statement here to
// minimize steps if energy is the same
// Now after this is done, you should have the cheapest cost to
// each vertex in "dist". Take the cheapest one on the edge.
// You can walk backwards on the "previous" to reconstruct the path taken.
这是一般的伪代码，虽然你也必须跟踪步骤的数量，主要是作为决胜局，所以它不应该是更多的工作。
对于DFS解决方案，它取决于能量是多少。如果它是有界的，小的和一个整数，你可以将二维图转换为x-y-e上的三维图形，其中e是剩余的能量 - 你从最初的能量开始，按照自己的方式工作向下，但要跟踪你以前去过的地方。
编辑：对于DFS解决方案，它应该是O(H*W*E)，对于E <= 30000，H，W <= 500，它可能不够快，至少是实时的，并且可能需要一点记忆。
答案 1 :(得分：4)
如果你害怕效率，就会采用A *方式：查看here。或者如果你觉得勇敢，你可以去D * here ..两者都会更有效率（因为他们使用启发式来估计距离）但不是很难实现！
BFS定义不是实现路径查找的好方法，它只是简单..
答案 2 :(得分：0)
这是一个实现，它使用您必须移动到达出口的路线对地图进行编码。它从整个地图中的任何可访问点找到最短路径：
import java.util.ArrayList;
import java.util.List;
public class Maze {
private static final char E = 'E'; // Ending position
private static final char X = 'X'; // Wall
private static final char O = ' '; // Space
private static final char L = 'L'; // Left
private static final char R = 'R'; // Right
private static final char U = 'U'; // Up
private static final char D = 'D'; // Down
private static final char FALSE = '0'; // Not accessible
private static final char TRUE = '1'; // Is accessible
private static final Node END_NODE = new Node(4, 4);
public static void main(String[] args) {
char[][] maze = new char[][] {
{X, X, X, X, X, X},
{X, O, O, X, O, X},
{X, O, X, X, O, X},
{X, O, O, O, X, X},
{X, X, O, X, O, X},
{X, O, O, O, O, X},
{X, O, X, X, O, X},
{X, X, X, X, X, X}};
// PLOT THE DESTINATION CELL AND ADD IT TO LIST
List<Node> nodes = new ArrayList<Node>();
nodes.add(END_NODE);
maze[END_NODE.row][END_NODE.col] = E;
// PRINT THE MAZE BEFORE ANY CALCULATIONS
printMaze(maze);
// SOLVE THE MAZE
fillMaze(maze, nodes);
printMaze(maze);
// CONVERT MAZE TO AN ADJACENCY MATRIX
compileMaze(maze);
printMaze(maze);
}
/**
* The parallel arrays define all four directions radiating from
* the dequeued node's location.
*
* Each node will have up to four neighboring cells; some of these
* cells are accessible, some are not.
*
* If a neighboring cell is accessible, we encode it with a directional
* code that calculates the direction we must take should we want to
* navigate to the dequeued node's location from this neighboring cell.
*
* Once encoded into our maze, this neighboring cell is itself queued
* up as a node so that recursively, we can encode the entire maze.
*/
public static final void fillMaze(char[][] maze, List<Node> nodes) {
int[] rowDirections = {-1, 1, 0, 0};
int[] colDirections = { 0, 0, -1, 1};
// dequeue our first node
Node destination = nodes.get(0);
nodes.remove(destination);
// examine all four neighboring cells for this dequeued node
for(int index = 0; index < rowDirections.length; index++) {
int rowIndex = destination.row + rowDirections[index];
int colIndex = destination.col + colDirections[index];
// if this neighboring cell is accessible, encode it and add it
// to the queue
if(maze[rowIndex][colIndex] == O) {
maze[rowIndex][colIndex] = getOppositeDirection(rowDirections[index], colDirections[index]);
nodes.add(new Node(rowIndex, colIndex));
}
}
// if our queue is not empty, call this method again recursively
// so we can fill entire maze with directional codes
if(nodes.size() > 0) {
fillMaze(maze, nodes);
}
}
/**
* Converts the maze to an adjacency matrix.
*/
private static void compileMaze(char[][] maze) {
for(int r = 0; r < maze.length; r++) {
for(int c = 0; c < maze[0].length; c++) {
if(maze[r][c] == X || maze[r][c] == O) {
maze[r][c] = FALSE;
}
else {
maze[r][c] = TRUE;
}
}
}
}
/**
* prints the specified two dimensional array
*/
private static final void printMaze(char[][] maze) {
System.out.println("====================================");
for(int r = 0; r < maze.length; r++) {
for(int c = 0; c < maze[0].length; c++) {
System.out.print(maze[r][c] + " ");
}
System.out.print("\n");
}
System.out.println("====================================");
}
/**
* Simply returns the opposite direction from those specified
* by our parallel direction arrays in method fillMaze.
*
* coordinate 1, 1 is the center of the char[][] array and
* applying the specified row and col offsets, we return the
* correct code (opposite direction)
*/
private static char getOppositeDirection(int row, int col) {
char[][] opposites = new char[][] {{O, D, O},{R, O, L},{O, U, O}};
return opposites[1 + row][1 + col];
}
}
class Node {
int row;
int col;
public Node(int rowIndex, int colIndex) {
row = rowIndex;
col = colIndex;
}
}
以下是这个小程序的输出
====================================
X X X X X X
X X X
X X X X
X X X
X X X E X
X X
X X X X
X X X X X X
====================================
====================================
X X X X X X
X D L X X
X D X X X
X R D L X X
X X D X E X
X R R R U X
X U X X U X
X X X X X X
====================================
====================================
0 0 0 0 0 0
0 1 1 0 0 0
0 1 0 0 0 0
0 1 1 1 0 0
0 0 1 0 1 0
0 1 1 1 1 0
0 1 0 0 1 0
0 0 0 0 0 0
====================================