Maze Problem with weight 1 and 2-优快云博客

本文介绍了一个模拟救援场景的问题：一名学生需要尽快从监狱中救出被囚禁的Hong。监狱地图由道路、障碍和墙壁组成，学生可以清除障碍来缩短救援时间。文章提供了问题描述、输入输出格式及示例，并附带了Java实现代码。

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Description

Hong was kidnapped by a bad woman. She was put in a prison. She transmitted distress signals to S students to save her as soon as possible.

The prison can be considered as a N\times MN×M matrix. Each grid can be road, barrier or wall. In each unit time, a student can move to an adjacent road(two grid are adjacent if and only if they have public side) or change an adjacent barrier to road. Wall can not be passed.

These S students are originally at S different positions and depart at t = 0 . Each of them wants to save Hong as soon as possible. Hong wants to know when she will be saved.

Input

The first line contains two integers N and M .

Each of the next N lines contains M chars, 'R' is a grid of road, 'B' is a grid of barrier, 'W' is a grid of wall, 'H' is the position of Hong, 'S' is the position of a student.

Output

Output a single integer, shows the minimal time Hong can be saved.

We guarantee that there must be a way to save Hong.

Sample Input

3 3
SBW
WRB
RBH

Copy

Sample output

Copy

Limit

1 second for each test case. The memory limit is 256MB.

For 20% test cases, N,M \leq 200N,M≤200.

For 50% test cases, there is no barrier.

For 100% test cases, N,M \leq 2000N,M≤2000.

Solution

import java.io.*;
import java.util.*;
class Node implements Comparable<Node>{
   int x, y, dist = 0;
   Node(int x, int y, int dist){this.x = x; this.y = y; this.dist = dist;}
   @Override
   public int compareTo(Node node){
	  return dist - node.dist;
   }
}
public class Main{
   static int[] dx = {-1, 1,0, 0 };
   static int[] dy = {0, 0, 1, -1};
   static int N = 0;
   static int M = 0;
   static char[][] state = null;
   public static void main(String[] args)throws Exception{
	  InputStream sys = System.in;
	 // InputStream fil = new FileInputStream(new File("C:\\Users\\ASUS\\desktop\\data4.txt"));
	  long b = System.currentTimeMillis();
	  BufferedReader in = new BufferedReader(new InputStreamReader(sys));
	  PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
	  StringTokenizer st;
	  st = new StringTokenizer(in.readLine()); //buffer: 31KiB
	  N = Short.valueOf(st.nextToken());
	  M = Short.valueOf(st.nextToken());
	  state = new char[N][M];//2000*2000*8 bits 30MiB W:0 R:1 S:2 B:3
	  int H_x = 0, H_y = 0;
	  char[] line;
	  for(short n = 0; n < N; ++n) {
		 line = in.readLine().toCharArray();
		 for(short m = 0; m < M; ++m) {
			if (line[m] == 'H') {
			   H_x = n;
			   H_y = m;
			   state[n][m] = 'R';
			}else{
			   state[n][m] = line[m];
			}
		 }
	  }
	  PriorityQueue<Node> queue = new PriorityQueue<Node>();
	  Node H = new Node(H_x, H_y, 0);
	  queue.add(H);
	  Node p = null;
	  boolean Found = false;
	  while(!queue.isEmpty() && !Found) {
		 p = queue.poll();
		 for(int i = 0; i < 4; ++i) {
			int x = p.x + dx[i];
			int y = p.y + dy[i];
			if (x < 0 || x >= N || y < 0 || y >= M) continue;
			if (state[x][y] == 'W') continue;
			if (state[x][y] == 'S') {
			   out.println(p.dist + 1);
			   Found = true;
			   break;
			}else if (state[x][y] == 'B') {
			   queue.add(new Node(x, y, p.dist + 2));
			}else {
			   queue.add(new Node(x, y, p.dist + 1));
			}
			state[x][y] = 'W';
		 }
	  }
	  //+" ("+D.x+","+D.y+")"
	  out.flush();
	 // System.out.println((System.currentTimeMillis() - b )*1.0/1000.0+"s");
   }
}