poj-1915-Knight Moves(广搜)
来源:互联网 发布:美工专业就业前景 编辑:程序博客网 时间:2024/05/06 01:20
Knight Moves
Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 19683 Accepted: 9084
Description
Background
Mr Somurolov, fabulous chess-gamer indeed, asserts that no one else but him can move knights from one position to another so fast. Can you beat him?
The Problem
Your task is to write a program to calculate the minimum number of moves needed for a knight to reach one point from another, so that you have the chance to be faster than Somurolov.
For people not familiar with chess, the possible knight moves are shown in Figure 1.
Mr Somurolov, fabulous chess-gamer indeed, asserts that no one else but him can move knights from one position to another so fast. Can you beat him?
The Problem
Your task is to write a program to calculate the minimum number of moves needed for a knight to reach one point from another, so that you have the chance to be faster than Somurolov.
For people not familiar with chess, the possible knight moves are shown in Figure 1.
Input
The input begins with the number n of scenarios on a single line by itself.
Next follow n scenarios. Each scenario consists of three lines containing integer numbers. The first line specifies the length l of a side of the chess board (4 <= l <= 300). The entire board has size l * l. The second and third line contain pair of integers {0, ..., l-1}*{0, ..., l-1} specifying the starting and ending position of the knight on the board. The integers are separated by a single blank. You can assume that the positions are valid positions on the chess board of that scenario.
Next follow n scenarios. Each scenario consists of three lines containing integer numbers. The first line specifies the length l of a side of the chess board (4 <= l <= 300). The entire board has size l * l. The second and third line contain pair of integers {0, ..., l-1}*{0, ..., l-1} specifying the starting and ending position of the knight on the board. The integers are separated by a single blank. You can assume that the positions are valid positions on the chess board of that scenario.
Output
For each scenario of the input you have to calculate the minimal amount of knight moves which are necessary to move from the starting point to the ending point. If starting point and ending point are equal,distance is zero. The distance must be written on a single line.
Sample Input
380 07 01000 030 50101 11 1
Sample Output
5280
Source
TUD Programming Contest 2001, Darmstadt, Germany
import java.util.ArrayDeque;import java.util.Queue;import java.util.Scanner;public class Main {/** * @param args */static Queue<Status> queue=new ArrayDeque<Status>();static boolean[][] mark;static int end_x;static int end_y;static int[][] dir={{-2,1},{-1,2},{1,2},{2,1},{2,-1},{1,-2},{-1,-2},{-2,-1}};static int n;public static void main(String[] args) {// TODO Auto-generated method stubScanner input=new Scanner(System.in);int m=input.nextInt();while(m-->0){queue.clear();n=input.nextInt();mark=new boolean[n][n];int start_x=input.nextInt();int start_y=input.nextInt();end_x=input.nextInt();end_y=input.nextInt();Status start=new Status();start.setX(start_x);start.setY(start_y);start.setStep(0);queue.add(start);int result=bfs(queue); System.out.println(result);}}//mainprivate static int bfs(Queue<Status> queue) {// TODO Auto-generated method stubwhile(!queue.isEmpty()){Status s=queue.poll();if(s.getX()==end_x && s.getY()==end_y)return s.step;for(int i=0;i<dir.length;++i){int x=s.getX()+dir[i][0];int y=s.getY()+dir[i][1];if(x<0 || x>=n || y<0 || y>=n)continue;Status status=new Status();status.setX(x);status.setY(y);status.setStep(s.step+1);if(mark[x][y]==false){queue.add(status);mark[x][y]=true;}}}return 0;}}class Status{int x;int y;int step;public int getX() {return x;}public void setX(int x) {this.x = x;}public int getY() {return y;}public void setY(int y) {this.y = y;}public int getStep() {return step;}public void setStep(int step) {this.step = step;}}
- poj 1915 Knight Moves(双向广搜)
- Poj 1915 - Knight Moves 双向广搜
- poj-1915-Knight Moves(广搜)
- POJ 1915 Knight Moves(广搜)
- 1915 Knight Moves 广搜
- POJ-1915 Knight Moves(双向广搜BFS)
- 简单广搜:Knight Moves
- poj 1915 Knight Moves
- poj 1915 Knight Moves
- poj 1915 Knight Moves
- POJ 1915 Knight Moves
- poj 1915 Knight Moves
- POJ 1915 Knight Moves
- POJ 1915 Knight Moves
- POJ 1915 Knight Moves
- POJ 1915 Knight Moves
- poj 1915 Knight Moves
- poj 1915 Knight Moves
- 2_回家
- HDUOJ 4681 2013多校第8场第6题 String
- 屌丝逆袭,你能付出多少
- hdu 1071 数学
- LINUX的文件分类及软硬链接文件
- poj-1915-Knight Moves(广搜)
- 探索Oracle11gR2 之 DataGuard_03 三种保护模式
- uva 10369
- Kendo UI开发教程(10): Kendo MVVM (一) 概述
- vb吐槽班04 一句话,不想多说
- hdu 2560(1.2.6)
- Cottage Village
- LeetCode-Scramble String
- Spark RDDs(弹性分布式数据集):为内存中的集群计算设计的容错抽象