• poj 2488 A Knight's Journey(dfs+字典序路径输出)


    转载请注明出处:http://blog.csdn.net/u012860063?viewmode=contents

    题目链接:http://poj.org/problem?

    id=2488


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    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

    3
    1 1
    2 3
    4 3

    Sample Output

    Scenario #1:
    A1
    
    Scenario #2:
    impossible
    
    Scenario #3:
    A1B3C1A2B4C2A3B1C3A4B2C4

    大致题意:

    给出一个国际棋盘的大小,推断马是否能不反复的走过全部格,并记录下当中按字典序排列的第一种路径。


    代码例如以下:

    #include <cstdio>
    #include <cstring>
    #define M 26
    struct node
    {
    	int x, y;
    }w[M*M];
    bool vis[M][M];
    int p, q;
    int flag = 0;
    int dir[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};
    //按此顺序搜索出来的结果就是字典序
    bool judge(int x, int y)
    {
    	if(x>=0&&x<q&&y>=0&&y<p&&!vis[x][y])
    		return true;
    	return false;
    }
    void dfs(int x, int y, int step)
    {
    	w[step].x = x,w[step].y = y;
    	vis[x][y] = true;
    	if(step == p*q-1)
    	{
    		flag = 1;
    		return ;
    	}
    	for(int i = 0; i < 8; i++)
    	{
    		int dx = w[step].x+dir[i][0];
    		int dy = w[step].y+dir[i][1];
    		if(judge(dx,dy))
    		{
    			vis[dx][dy] = true;
    			dfs(dx,dy,step+1);
    			if(flag)//一但找到就退出搜索
    				return;
    			vis[dx][dy] = false;
    		}
    	}
    	return ;
    }
    void print()
    {
    	for(int i = 0; i < p*q; i++)//列为字母,行为数字
    	{
    		printf("%c%d",w[i].x+'A',w[i].y+1);
    	}
    	printf("
    
    ");
    }
    int main()
    {
    	int t, i, j, cas = 0;
    	scanf("%d",&t);
    		while(t--)
    		{
    			memset(vis,false,sizeof(vis));
    			flag = 0;
    			scanf("%d%d",&p,&q);
    			for(i = 0; i < q; i++)//列
    			{
    				for(j = 0; j < p; j++)//行
    				{
    					dfs(i,j,0);
    					if(flag)
    						break;
    				}
    				if(flag)
    					break;
    			}
    			printf("Scenario #%d:
    ",++cas);
    			if(flag)
    				print();
    			else
    				printf("impossible
    
    ");
    		}
    	return 0;
    }



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  • 原文地址:https://www.cnblogs.com/yangykaifa/p/6991557.html
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