1、题目
矩阵中有3种类型：0仓库，-1障碍，1零售店。现在每个零售店要去距离它最近的仓库取货物，请计算出所有零售店到最近仓库距离之和，假设矩阵中每个单元格之间距离为1。如果遇到障碍物，则表示无法通过。可以在单元格上下左右移动。
注意
无法到达仓库的零售店不参与计算
没有零售店或者没有仓库，则返回0
2、用例
第一行输入3 3代表3×3矩阵
第二行输入矩阵值
第三行输出最短距离和
3 3
1 -1 0
0 1 1
1 -1 1
结果为6
3、算法
BFS
public class BFS {// 0仓库，-1障碍，1零售店// 无法到达仓库的零售店不参与计算// 没有零售店或者没有仓库，则返回0public static void main(String[] args) {Scanner cin = new Scanner(System.in, StandardCharsets.UTF_8.name());int rows = cin.nextInt();int cols = cin.nextInt();int[][] grids = new int[rows][cols];for (int i = 0; i < rows; i++) {for (int j = 0; j < cols; j++) {grids[i][j] = cin.nextInt();}}cin.close();System.out.println(nearestWareHouse(grids));}private static int nearestWareHouse(int[][] grid) {int rows = grid.length;int cols = grid[0].length;// 仓库节点坐标Queue<int[]> visitQueue = new LinkedList<>();// 遍历grid，把所有仓库节点坐标放入bfs队列for (int i = 0; i < rows; i++) {for (int j = 0; j < cols; j++) {if (grid[i][j] == 0) {// 如果是仓库节点，入队，设置visited[i][j] = truevisitQueue.add(new int[]{i, j});}}}// 初始化步长为1，此时队列中为仓库，以仓库为中心，以1为步长，向四周扩散// 将已遍历的节点设置为-1int step = 1;int result = 0;while (!visitQueue.isEmpty()) {int qcount = visitQueue.size();for (int i = 0; i < qcount; i++) {// 出队列，向该节点四周扩散// poll从队列中移除并返回头部元素int[] temp = visitQueue.poll();int x = temp[0];int y = temp[1];// 右边if (x + 1 < rows && grid[x + 1][y]==1) {result += step;grid[x + 1][y] = -1;visitQueue.add(new int[]{x + 1, y});}// 左边if (x - 1 >= 0 && grid[x - 1][y]==1) {result += step;grid[x - 1][y] = -1;visitQueue.add(new int[]{x - 1, y});}// 上面if (y - 1 >= 0 && grid[x][y - 1]==1) {result += step;grid[x][y - 1] = -1;visitQueue.add(new int[]{x, y - 1});}// 下面if (y + 1 < cols && grid[x][y + 1]==1) {result += step;grid[x][y + 1] = -1;visitQueue.add(new int[]{x, y + 1});}}step++;}return result;}
}
BFS2
static class Point {int x;int y;int step;public Point(int x, int y, int step) {this.x = x;this.y = y;this.step = step;}}private static int nearestWareHouse(int[][] grid) {int result = 0;int rows = grid.length;int cols = grid[0].length;Queue<Point> queue = new LinkedList<>();// 遍历grid，把所有仓库节点坐标放入bfs队列for (int i = 0; i < rows; i++) {for (int j = 0; j < cols; j++) {if (grid[i][j] == 0) {// 如果是仓库节点，入队queue.add(new Point(i, j, 0));}}}while (!queue.isEmpty()) {Point point = queue.poll();int x = point.x;int y = point.y;// 左if (x != 0 && grid[x - 1][y] == 1) {queue.add(new Point(x - 1, y, point.step + 1));grid[x - 1][y] = -1;}// 右if (x != rows - 1 && grid[x + 1][y] == 1) {queue.add(new Point(x + 1, y, point.step + 1));grid[x + 1][y] = -1;}// 上if (y != 0 && grid[x][y - 1] == 1) {queue.add(new Point(x + 1, y, point.step + 1));grid[x + 1][y] = -1;}// 下if (y != cols - 1 && grid[x][y + 1] == 1) {queue.add(new Point(x, y + 1, point.step + 1));grid[x][y + 1] = -1;}result += point.step;}return result;}
BFS3
private static int nearestWareHouse(int[][] grid) {int rows = grid.length;int cols = grid[0].length;// 仓库节点坐标Queue<int[]> visitQueue = new LinkedList<>();// 记录已访问和未访问标记boolean[][] visited = new boolean[rows][cols];// 遍历grid，把所有仓库节点坐标放入bfs队列for (int i = 0; i < rows; i++) {for (int j = 0; j < cols; j++) {if (grid[i][j] == 0) {// 如果是仓库节点，入队，设置visited[i][j] = true，设置未访问过visitQueue.add(new int[]{i, j});visited[i][j] = false;} else if (grid[i][j] == -1) {// 障碍物，设置未访问过visited[i][j] = false;} else {// 仓库，设置已访问过visited[i][j] = true;}}}// 初始化步长为1，此时队列中为仓库，以仓库为中心，以1为步长，向四周扩散// 将已遍历的节点设置为-1int step = 1;int result = 0;while (!visitQueue.isEmpty()) {int qcount = visitQueue.size();for (int i = 0; i < qcount; i++) {// 出队列，向该节点四周扩散// poll从队列中移除并返回头部元素int[] temp = visitQueue.poll();int x = temp[0];int y = temp[1];// 右边if (x + 1 < rows && visited[x + 1][y]) {result += step;visited[x + 1][y] = false;visitQueue.add(new int[]{x + 1, y});}// 左边if (x - 1 >= 0 && visited[x - 1][y]) {result += step;visited[x - 1][y] = false;visitQueue.add(new int[]{x - 1, y});}// 上面if (y - 1 >= 0 && visited[x][y - 1]) {result += step;visited[x][y - 1] = false;visitQueue.add(new int[]{x, y - 1});}// 下面if (y + 1 < cols && visited[x][y + 1]) {result += step;visited[x][y + 1] = false;visitQueue.add(new int[]{x, y + 1});}}step++;}return result;}
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