HDU 1026 Ignatius and the Princess I （优先队列+BFS（广度优先搜索））

Ignatius and the Princess I

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 10051 Accepted Submission(s): 3023

Special Judge

Problem Description

The Princess has been abducted by the BEelzebub feng5166, our hero Ignatius has to rescue our pretty Princess. Now he gets into feng5166's castle. The castle is a large labyrinth. To make the problem simply, we assume the labyrinth is a N*M two-dimensional array which left-top corner is (0,0) and right-bottom corner is (N-1,M-1). Ignatius enters at (0,0), and the door to feng5166's room is at (N-1,M-1), that is our target. There are some monsters in the castle, if Ignatius meet them, he has to kill them. Here is some rules:

1.Ignatius can only move in four directions(up, down, left, right), one step per second. A step is defined as follow: if current position is (x,y), after a step, Ignatius can only stand on (x-1,y), (x+1,y), (x,y-1) or (x,y+1).
2.The array is marked with some characters and numbers. We define them like this:
. : The place where Ignatius can walk on.
X : The place is a trap, Ignatius should not walk on it.
n : Here is a monster with n HP(1<=n<=9), if Ignatius walk on it, it takes him n seconds to kill the monster.

Your task is to give out the path which costs minimum seconds for Ignatius to reach target position. You may assume that the start position and the target position will never be a trap, and there will never be a monster at the start position.

 

Input

The input contains several test cases. Each test case starts with a line contains two numbers N and M(2<=N<=100,2<=M<=100) which indicate the size of the labyrinth. Then a N*M two-dimensional array follows, which describe the whole labyrinth. The input is terminated by the end of file. More details in the Sample Input.

 

Output

For each test case, you should output "God please help our poor hero." if Ignatius can't reach the target position, or you should output "It takes n seconds to reach the target position, let me show you the way."(n is the minimum seconds), and tell our hero the whole path. Output a line contains "FINISH" after each test case. If there are more than one path, any one is OK in this problem. More details in the Sample Output.

 

Sample Input

5 6 .XX.1. ..X.2. 2...X. ...XX. XXXXX. 5 6 .XX.1. ..X.2. 2...X. ...XX. XXXXX1 5 6 .XX... ..XX1. 2...X. ...XX. XXXXX.

 

Sample Output

It takes 13 seconds to reach the target position, let me show you the way. 1s:(0,0)->(1,0) 2s:(1,0)->(1,1) 3s:(1,1)->(2,1) 4s:(2,1)->(2,2) 5s:(2,2)->(2,3) 6s:(2,3)->(1,3) 7s:(1,3)->(1,4) 8s:FIGHT AT (1,4) 9s:FIGHT AT (1,4) 10s:(1,4)->(1,5) 11s:(1,5)->(2,5) 12s:(2,5)->(3,5) 13s:(3,5)->(4,5) FINISH It takes 14 seconds to reach the target position, let me show you the way. 1s:(0,0)->(1,0) 2s:(1,0)->(1,1) 3s:(1,1)->(2,1) 4s:(2,1)->(2,2) 5s:(2,2)->(2,3) 6s:(2,3)->(1,3) 7s:(1,3)->(1,4) 8s:FIGHT AT (1,4) 9s:FIGHT AT (1,4) 10s:(1,4)->(1,5) 11s:(1,5)->(2,5) 12s:(2,5)->(3,5) 13s:(3,5)->(4,5) 14s:FIGHT AT (4,5) FINISH God please help our poor hero. FINISH

 

题意：一个N*M的图，一秒内可上下左右选个方向走一步，一进去，若有数字，就要花数字大小的时间停留在那个格子，输出从起点(0, 0)到终点(N-1, M-1)每一秒的路径。

思路： 优先队列+BFS（广度优先搜索）

import java.io.*;
import java.util.*;
public class Main {
	int n,m;
	char ch[][];
	boolean boo[][];
	int fx[]={1,-1,0,0};
	int fy[]={0,0,-1,1};
	Queue<Node> que=new PriorityQueue<Node>();//优先队列
	public static void main(String[] args) {
		new Main().work();
	}
	void work(){
		Scanner sc=new Scanner(new BufferedInputStream(System.in));
		while(sc.hasNext()){
			que.clear();
			n=sc.nextInt();
			m=sc.nextInt();
			ch=new char[n][m];
			boo=new boolean[n][m];
			for(int i=0;i<n;i++){
				String s=sc.next();
				ch[i]=s.toCharArray();
				Arrays.fill(boo[i],false);
			}
			Node node=new Node();
			node.x=0;
			node.y=0;
			node.t=0;
			String s1="("+node.x+","+node.y+")";
			node.s=s1+" ";
			boo[node.x][node.y]=true;;
			que.add(node);
			BFS();
		}
	}
	void BFS(){
		while(!que.isEmpty()){
			Node node=que.poll();
			if((node.x==n-1)&&(node.y==m-1)){
				out(node.s,node.t);
				return;
			}
			for(int i=0;i<4;i++){
				int px=node.x+fx[i];
				int py=node.y+fy[i];
				if(check(px,py)&&!boo[px][py]){
					boo[px][py]=true;
					Node td=node.getNode();
					td.x=px;
					td.y=py;
					String s1="("+td.x+","+td.y+")";
					td.s+=s1+" ";
					if(ch[px][py]=='.'){
						td.t=td.t+1;
					}
					else{
						td.t=td.t+(ch[px][py]-'0');
						td.t=td.t+1;
						for(int j=0;j<ch[px][py]-'0';j++){
							td.s+=s1+" ";
						}
					}
					que.add(td);
				}
			}
		}
		System.out.println("God please help our poor hero.");
		System.out.println("FINISH");
	}
	//输出
	void out(String s,int time){
		System.out.println("It takes "+time+" seconds to reach the target position, let me show you the way.");
		String str[]=s.split(" ");
		for(int i=1;i<str.length;i++){
			if(str[i].length()!=0){
				System.out.print(i+"s:");
				if(str[i-1].equals(str[i])){
					System.out.println("FIGHT AT "+str[i]);
				}
				else{
					System.out.println(str[i-1]+"->"+str[i]);
				}
			}
		}
		System.out.println("FINISH");
	}
	boolean check(int px,int py){
		if(px<0||px>n-1||py<0||py>m-1||ch[px][py]=='X')
			return false;
		return true;
	}
	class Node implements Comparable<Node>{
		int x;
		int y;
		int t;
		String s;
		public int getT() {
			return t;
		}
		public void setT(int t) {
			this.t = t;
		}
		Node getNode(){
			Node node=new Node();
			node.t=t;
			node.s=s;
			return node;
		}
		public int compareTo(Node o) {//按升序排序
			return this.t>o.t?1:-1;
		}
	}
}