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package class_2022_10_2_week;
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// 定义一个二维数组N*M，比如5*5数组下所示：
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// 0, 1, 0, 0, 0,
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// 0, 1, 1, 1, 0,
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// 0, 0, 0, 0, 0,
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// 0, 1, 1, 1, 0,
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// 0, 0, 0, 1, 0,
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// 它表示一个迷宫，其中的1表示墙壁，0表示可以走的路
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// 只能横着走或竖着走，不能斜着走
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// 要求编程序找出从左上角到右下角距离最短的路线
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// 测试链接 : https://www.nowcoder.com/practice/cf24906056f4488c9ddb132f317e03bc
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// 请同学们务必参考如下代码中关于输入、输出的处理
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// 这是输入输出处理效率很高的写法
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// 提交以下的code，提交时请把主类名改成"Main"
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import java.io.BufferedReader;
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import java.io.IOException;
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import java.io.InputStreamReader;
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import java.io.OutputStreamWriter;
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import java.io.PrintWriter;
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import java.io.StreamTokenizer;
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import java.util.ArrayList;
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import java.util.PriorityQueue;
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public class Code03_FindMazeMinPath {
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public static void main(String[] args) throws IOException {
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BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
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StreamTokenizer in = new StreamTokenizer(br);
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PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
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while (in.nextToken() != StreamTokenizer.TT_EOF) {
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int n = (int) in.nval;
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in.nextToken();
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int m = (int) in.nval;
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int[][] map = new int[n][m];
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < m; j++) {
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in.nextToken();
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map[i][j] = (int) in.nval;
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}
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}
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ArrayList<int[]> ans = dijkstra(n, m, map);
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for (int i = ans.size() - 1; i >= 0; i--) {
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out.println("(" + ans.get(i)[0] + "," + ans.get(i)[1] + ")");
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}
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out.flush();
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}
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}
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// n : n行
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// m : m列
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// map :
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// 0 1 1 1
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// 0 0 0 1
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// 1 1 0 1
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// 0 0 0 0
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// list = [0,0] , [1,0], [1,1]...[3,3]
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// [3,3] -> [0,0]
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public static ArrayList<int[]> dijkstra(int n, int m, int[][] map) {
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// (a,b) -> (c,d)
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// last[c][d][0] = a
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// last[c][d][1] = b
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// 从哪到的当前(c,d)
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int[][][] last = new int[n][m][2];
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// int[] arr = {c,d,w}
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// 0 1 距离
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PriorityQueue<int[]> heap = new PriorityQueue<>((a, b) -> a[2] - b[2]);
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boolean[][] visited = new boolean[n][m];
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heap.add(new int[] { 0, 0, 0 });
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ArrayList<int[]> ans = new ArrayList<>();
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while (!heap.isEmpty()) {
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int[] cur = heap.poll();
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int x = cur[0];
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int y = cur[1];
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int w = cur[2];
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if (x == n - 1 && y == m - 1) {
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break;
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}
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if (visited[x][y]) {
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continue;
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}
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// (x,y)这个点
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visited[x][y] = true;
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add(x, y, x - 1, y, w, n, m, map, visited, heap, last);
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add(x, y, x + 1, y, w, n, m, map, visited, heap, last);
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add(x, y, x, y - 1, w, n, m, map, visited, heap, last);
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add(x, y, x, y + 1, w, n, m, map, visited, heap, last);
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}
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int x = n - 1;
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int y = m - 1;
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while (x != 0 || y != 0) {
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ans.add(new int[] { x, y });
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int lastX = last[x][y][0];
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int lastY = last[x][y][1];
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x = lastX;
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y = lastY;
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}
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ans.add(new int[] { 0, 0 });
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return ans;
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}
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// 当前是从(x,y) -> (i,j)
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// 左上角 -> (x,y) , 距离是w
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// 左上角 -> (x,y) -> (i,j) w+1
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// 行一共有n行，0~n-1有效
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// 列一共有m行，0~m-1有效
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// map[i][j] == 1，不能走！是障碍！
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// map[i][j] == 0，能走！是路！
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// 把记录加入到堆里，所以得有heap
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// last[i][j][0] = x
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// last[i][j][1] = y
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public static void add(int x, int y,
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int i, int j, int w,
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int n, int m,
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int[][] map,
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boolean[][] visited,
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PriorityQueue<int[]> heap,
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int[][][] last) {
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if (i >= 0 && i < n
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&& j >= 0 && j < m
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&& map[i][j] == 0
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&& !visited[i][j]) {
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heap.add(new int[] { i, j, w + 1 });
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last[i][j][0] = x;
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last[i][j][1] = y;
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}
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}
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} |
