POJ-2488-A Knight's Journey

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 40833		Accepted: 13882

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
// 马的走法很恶心,横坐标按从小到大的顺序,横坐标相同纵坐标从小到大
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
int n, m, t;
int Map[100][100];
int tx[]= { -2,-2,-1,  -1,1,1,2,2};
int ty[]= { -1,1,-2,2,-2,2,-1,1};
int flag;
int p,top;
struct node
{
int x;
int y;
}way[10010];
void dfs(int x, int y,int s)
{
if(s==n*m)
{
flag=1;
return ;
}
for(int i = 0; i<8; ++i)
{
int xx =x+tx[i];
int yy =y+ty[i];
if(xx>=1&&xx<=n&&yy>=1&&yy<=m&&!Map[xx][yy])
{
Map[xx][yy] = 1;
way[top].x = xx;
way[top].y = yy;
++top;
dfs(xx, yy, s+1);
if(flag)return;
--top;
Map[xx][yy] = 0;
}
}

}
int main()
{

scanf("%d", &t);
for(int tt = 1; tt<=t ;++tt)
{
scanf("%d%d", &m, &n);
memset(Map, 0,sizeof(Map));
printf("Scenario #%d:\n",tt);
flag = 0;
top = 1;
way[0].x = 1;
way[0].y = 1;
Map[1][1] = 1;
dfs(1,1,1);
if(!flag)printf("impossible\n");
else
{
for(int i = 0; i<top;++i)
printf("%c%d",way[i].x+'A'-1,way[i].y);
printf("\n");
}
printf("\n");

}

return 0;
}