A Knight's Journey


Time Limit: 1000MS

 

Memory Limit: 65536K

Total Submissions: 37150

 

Accepted: 12594


Description




POJ2488（A Knight

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 1 2 3 4 3


Sample Output


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




题意：判断一个骑士是否能走完整个棋盘上的点，按国际象棋中马的规则走，输出字典序最小的序列~



国际象棋盘上点的横坐标是A,B,C,D。。。纵坐标是1，2，3，4。。。


既然要走完所有的点，并且保证当前要走的点是字典序最小的点。所以每次都要从左上角那点开始走


用int s[8][2]={-2,-1,-2,1,-1,-2,-1,2,1,-2,1,2,2,-1,2,1}; 来实现搜索过程，并且


保证输出的是字典序最小的序列~ 搜索到的第一路径就是骑士所走的路径。


用path[][0]用来存储路径的横坐标，path[][1]来存储路径的纵坐标。



#include<stdio.h>
#include<string.h>
int vis[30][30];
int path[30][2];
int s[8][2]={-2,-1,-2,1,-1,-2,-1,2,1,-2,1,2,2,-1,2,1};
int a,b,flag;
void dfs(int x,int y,int ans)
{
  if(ans==a*b)
  {
    flag=1;
    for(int i=0;i<ans;i++)
    {
      printf("%c%d",path[i][0]+'A',path[i][1]+1);
    }
    printf("\n");
  }
  for(int i=0;i<8;i++)
  {
    int m=x+s[i][0];
    int n=y+s[i][1];
    if(m>=0&&m<b&&n>=0&&n<a&&!flag&&!vis[m][n])
    {
      vis[m][n]=1;
      path[ans][0]=m;
      path[ans][1]=n;
      dfs(m,n,ans+1);
      vis[m][n]=0;
    }
  }
}
int main()
{
  int test,kase=1;
  scanf("%d",&test);
  while(test--)
  {
    scanf("%d%d",&a,&b);
    memset(vis,0,sizeof(vis));
    memset(path,0,sizeof(path));
    flag=0;
    vis[0][0]=1;
    printf("Scenario #%d:\n",kase++);
    dfs(0,0,1);
    if(!flag) printf("impossible\n");
    printf("\n");
  }
 }




用java语言实现上述过程，思想和代码基本上一样，就是换一下输入输出和一些格式~



import java.util.Scanner;
public class Main {

  private static int[][] a = { { -2, -1 }, 
                  { -2, 1 }, 
                  { -1, -2 }, 
                  { -1, 2 }, 
                  { 1, -2 }, 
                  { 1, 2 }, 
                  { 2, -1 },
                  { 2, 1 } }; 

  private static int[][] hang=new int[30][2];
  private static Boolean[][] visit=new Boolean[30][30];
  private static int x,y;  
  private static Boolean flag;


  public static void main(String[] args) {
    
    
    Scanner s = new Scanner(System.in);
    int num=s.nextInt();
    int key=1;
    while (num-->0) {
        flag=false;
         x=s.nextInt();
         y=s.nextInt();
         System.out.println("Scenario #"+key+++":");
         for (int i = 0; i < 30; i++) {
          for (int j = 0; j < 30; j++) {
            visit[i][j]=false;
          }
        }
            hang[0][0]=0;
        hang[0][1]=0;
        visit[0][0]=true;
        dfs(0,0,1);
        if (!flag) {
          System.out.println("impossible");
        }
        //System.out.println();
        if(num!=0){
          System.out.println();
        }
    }

  }  
  public static void dfs(int m,int n,int step){
    
    if(step==x*y){
      flag=true;
      for(int i=0;i<step;i++){
        System.out.print((char)(hang[i][0]+'A'));
        System.out.print(hang[i][1]+1);
      }
      System.out.println();
    }
    
    for(int i=0;i<8;i++){
        
      int h=m+a[i][0];
      int l=n+a[i][1];
      if(h<y&&h>=0&&l>=0&&l<x&&!flag&&!visit[h][l]){
        hang[step][0]=h;
        hang[step][1]=l;
        visit[h][l]=true;
        dfs(h,l,step+1);
        visit[h][l]=false;
      }
      
    }
    
  }
  
}