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POJ2488:A Knight's Journey(dfs)

2024-07-23 11:00:24   215人阅读

http://poj.org/problem?id=2488

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

31 12 34 3

Sample Output

Scenario #1:A1Scenario #2:impossibleScenario #3:A1B3C1A2B4C2A3B1C3A4B2C4

题目大意：

              给出国际象棋的的一个棋盘，里面有一个马，给出所在棋盘的长和宽，问从任意一点开始，能否不重复的走遍棋盘上的每一点，若能就输出一个字符串，其中字母代表行，数字代表列（按字典序搜索），不能就输出impossible

解题思路：

              很明显可以用DFS暴力搜索，但是要注意的是，骑士在搜索的时候按字典序的方向搜索。

题目解析：
一直没读懂题，做出来的答案一直和样例不同，后来看题解，说是要按字典序搜索，然后又是N遍WA，（只能说dfs自己学的很渣，递归一会就
递晕了），然后没敢深入思考，别人的代码，都是指dfs了一遍，即dfs(1,1,1),我不知道为什么，只好枚举所有结果，综合来说，算是一道
水题吧，只要知道按字典序搜索就好了。

#include <stdio.h>#include <iostream>#include <string.h>#include <stdlib.h>using namespace std;int jx[8]= {-2,-2,-1,-1,1,1,2,2};int jy[8]= {-1,1,-2,2,-2,2,-1,1};int m,n,v[28][28],flag;char f[28*28];int g[28*28];void pu(){    for(int i=1; i<=n*m; i++)    {        printf("%c%d",f[i],g[i]);    }    printf("\n");}void dfs(int x,int y,int ans){    int tx,ty;    if(ans==n*m)    {        f[ans]=x+‘A‘-1;        g[ans]=y;        flag=1;        return ;    }    for(int i=0; i<8; i++)    {        tx=x+jx[i];        ty=y+jy[i];        if(tx>=1&&tx<=n&&ty>=1&&ty<=m&&v[tx][ty]==0)        {            v[tx][ty]=1;            f[ans+1]=tx+‘A‘-1;            g[ans+1]=ty;            dfs(tx,ty,ans+1);            if(flag==1)                return ;            v[tx][ty]=0;        }    }    return ;}int main(){    int T;    scanf("%d",&T);    for(int z=1; z<=T; z++)    {        flag=0;        scanf("%d%d",&m,&n);        printf("Scenario #%d:\n",z);        for(int i=1; i<=n; i++)        {            for(int j=1; j<=m; j++)            {                memset(v,0,sizeof(v));                v[i][j]=1;                f[1]=i-1+‘A‘;                g[1]=j;                dfs(i,j,1);                if(flag==1)                {                    pu();                    break;                }            }            if(flag==1) break;        }        if(flag==0) printf("impossible\n");        printf("\n");    }    return 0;}

 

POJ2488:A Knight's Journey(dfs)


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