POJ 2488 A Knight's Journey

A Knight's Journey

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 25252 Accepted: 8584

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

Source

TUD Programming Contest 2005, Darmstadt, Germany

 

题意：

给出国际象棋棋盘的行数和列数，判断马能不能不重复地走遍所有格子，若能，则按字典序输出走动路径

在国际象棋中马走“日”字格

 

代码：

#include<cstdio>#include<cstring>#include<cstdlib>#include<string>#define N 27typedef struct Node{int row,col;}node;node way[N*N];int p,q;          //p为数字行，q为字母列bool chess[N][N]; //每次查找时标记该位置有没有被走到过int dx[8]={-1,1,-2,2,-2,2,-1,1};    //dfs时棋子走动顺序，注意顺序int dy[8]={-2,-2,-1,-1,1,1,2,2};bool dfs(int i,int j,int step)    //step为当前行走步数{chess[i][j]=true;way[step].row=i;way[step].col=j;if(step==p*q) return true;   //step==p*q则所有格子都被走到for(int k=0;k<8;k++){int ii=i+dx[k];int jj=j+dy[k];if(!chess[ii][jj] && ii>=1 && ii<=p && jj>='A' && jj<='A'+q-1)if(dfs(ii,jj,step+1))return true;}chess[i][j]=false;    //执行至此，前面8步均不能成立，返回上一层return false;}int main(){int Case=1,t;while(scanf("%d",&t)!=EOF){if(t==-1) break;while(t--){memset(chess,false,sizeof(chess));scanf("%d%d",&p,&q);bool flag=false;for(int j='A';j<='A'+q-1;j++){for(int i=1;i<=p;i++){memset(way,0,sizeof(way));if(dfs(i,j,1)){printf("Scenario #%d:\n",Case++);for(int k=1;k<=p*q;k++)printf("%c%d",way[k].col,way[k].row);printf("\n\n");flag=true;break;}}if(flag) break;}if(!flag) {printf("Scenario #%d:\n",Case++);printf("impossible\n\n");}}}return 0;}

思路：

因为要求按字典序输出，则要注意深度搜索时的顺序