A Knight's Journey-优快云博客

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 43095		Accepted: 14619

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

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<stdlib.h>
#include<time.h>
#include<math.h>
#include<queue>;
#include<stack>;
#include <iomanip>;
using namespace std;
int t,b[100][2],v[10][10],n,m,p;
int dir[8][2]= {{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}};
int go(int x,int y)
{
    if(1<=x&&x<=n&&1<=y&&y<=m&&!v[x][y]&&!p)
        return 1;
    return 0;
}
void dfs(int x,int y,int c)
{
    b[c][0]=y;
    b[c][1]=x;
    if(c==n*m)
    {
        p=1;
        return;
    }
    for(int i=0; i<8; i++)
    {
        int ex=x+dir[i][0];
        int ey=y+dir[i][1];
        if(go(ex,ey))
        {
            v[ex][ey]=1;
            dfs(ex,ey,c+1);
            v[ex][ey]=0;
        }
    }
}
int main()
{
    scanf("%d",&t);
    for(int i=1; i<=t; i++)
    {
         p=0;
        scanf("%d%d",&n,&m);
        memset(v,0,sizeof(v));
        printf("Scenario #%d:\n",i);
        v[1][1]=1;
        dfs(1,1,1);
        if(p==1)
        {
            for(int i=1; i<=n*m; i++)
            {
                printf("%c%d",b[i][0]+'A'-1,b[i][1]);
            }
        }
        else
            printf("impossible");
        printf("\n");
        if(i!=t)
            printf("\n");
    }
}