POJ 2488 A Knight's Journey

最新推荐文章于 2022-05-18 14:38:09 发布

原创最新推荐文章于 2022-05-18 14:38:09 发布 · 773 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#acm #C++ #algorithm #poj #oj

【OJ】POJ 同时被 2 个专栏收录

37 篇文章

订阅专栏

【Algorithm】DFS

16 篇文章

订阅专栏

本文介绍了一种寻找骑士在特定大小的棋盘上遍历所有格子的最短路径的方法，通过深度优先搜索（DFS）算法实现了路径规划，并确保了路径的字典序最小。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

来源:http://poj.org/problem?id=2488

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30588		Accepted: 10479

Description

Background
The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey
around the world. Whenever a knight moves, it is two squares in one direction and one square perpendicular to this. The world of a knight is the chessboard he is living on. Our knight lives on a chessboard that has a smaller area than a regular 8 * 8 board, but it is still rectangular. Can you help this adventurous knight to make travel plans?

Problem
Find a path such that the knight visits every square once. The knight can start and end on any square of the board.

Input

The input begins with a positive integer n in the first line. The following lines contain n test cases. Each test case consists of a single line with two positive integers p and q, such that 1 <= p * q <= 26. This represents a p * q chessboard, where p describes how many different square numbers 1, . . . , p exist, q describes how many different square letters exist. These are the first q letters of the Latin alphabet: A, . . .

Output

The output for every scenario begins with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the lexicographically first path that visits all squares of the chessboard with knight moves followed by an empty line. The path should be given on a single line by concatenating the names of the visited squares. Each square name consists of a capital letter followed by a number.
If no such path exist, you should output impossible on a single line.

Sample Input

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

Source

TUD Programming Contest 2005, Darmstadt, Germany

题意: 输出骑士遍历棋盘的路径，起点终点任意~ 要求路径的字典序要最小。如果无法遍历则输出impossible

题解: DFS 这里需要注意的字典序输出路径，开始我没查 lexicographically 是什么意思第三组样例就过不了~~这里实现字典序，只要搜索顺序稍微注意下就好

int dir[8][2]={
{-1,-2},{1,-2},
{-2,-1}, {2,-1},
{-2,1},{2,1},
{-1,2},{1,2}
};

这样就可以保证字典序了~~

PS: 后来我还是WA了，错误发现后很无语 impossible打成 Impossible ~~以后坚决不手打！

//要字典序输出 
#include<cstdio>
#include<cstring>
const int Max=30;
struct Node{
	bool square;
	int prex,prey;
} map[Max][Max];
int t,x,y,endx,endy,squares,pos,posx[Max][2];
int dir[8][2]={
	{-1,-2},{1,-2},
	{-2,-1}, {2,-1},
	 {-2,1},{2,1},
	{-1,2},{1,2}
};
bool flag=false; 
void dfs(int dx,int dy,int step){
	if(step==x*y){
		endx=dx; endy=dy;
		flag=true;
		return;
	}
	for(int i=0;i<8&&!flag;i++){
		int tempx=dx+dir[i][0],tempy=dy+dir[i][1];
	if(tempx>=0&&tempx<x&&tempy>=0&&tempy<y&&map[tempx][tempy].square){
		map[tempx][tempy].square=false;
		map[tempx][tempy].prex=dx;map[tempx][tempy].prey=dy;
		dfs(tempx,tempy,step+1);
		map[tempx][tempy].square=true;
	}
	}		
}
void  init(){
	for(int i=0;i<x;i++)
	for(int j=0;j<y;j++)
	map[i][j].square=true;
}
int main(){
	scanf("%d",&t);
	for(int k=1;k<=t;k++){
	scanf("%d%d",&x,&y);  squares=x*y;flag=false; 
	printf("Scenario #%d:\n",k);
	for(int i=0;i<x&&!flag;i++)
	for(int j=0;j<y&&!flag;j++){
		init( );      //初始化棋盘 
		map[i][j].prex=map[i][j].prey=-1;map[i][j].square=false;
		dfs(i,j,1);
		if(flag){  //成功 
		int px=endx,py=endy; pos=0;
		posx[pos][0]=px;posx[pos++][1]=py;
		//printf("%c%d",endx+'A',endy+1);
		while(map[px][py].prex!=-1&&map[px][py].prey!=-1){
		int tempx=px;
		px=map[px][py].prex;py=map[tempx][py].prey;
		posx[pos][0]=px;posx[pos++][1]=py;
		}
		for(int z=pos-1;z>=0;z--)
		printf("%c%d",posx[z][1]+'A',posx[z][0]+1);
		break;		
		}
	}
	if(!flag){//失败 
		printf("impossible");	
	} 
	printf("\n\n");
	}
	return 0;
}