A Knight's Journey
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 34633
Accepted: 11815
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
题目简单翻译：
给你一个象棋中的马，一个n*m的棋盘，求是否能从一点出发，走遍整个棋盘，不重复走。如果能，按字典序输出第一个序列。如果不能，则输出“impossible”。
解题思路：
dfs,从一点出发，然而因为要字典序较小的，我们就选择（1，1）为起始点吧。注意延伸的方向，优先向字典序小的方向延伸。
代码：
#include<cstdio>
#include<cstring>
#include<queue> using namespace std;
int n,m;
int vis[][];
int dx[]={-,-,-,-,,,,};
int dy[]={-,,-,,-,,-,};
int a1[],a2[];
bool check(int x,int y)
{
return x>=&&x<n&&y>=&&y<m;
}
bool dfs(int x,int y,int depth)
{
if(depth==m*n)
{
for(int i=;i<depth;i++)
printf("%c%d",a1[i]+'A',a2[i]+);
puts("");
return true;
}
for(int i=;i<;i++)
{
int curx=x+dx[i];
int cury=y+dy[i];
if(check(curx,cury)&&vis[curx][cury]==)
{
a1[depth]=curx;
a2[depth]=cury;
vis[curx][cury]=;
if(dfs(curx,cury,depth+)) return true;
vis[curx][cury]=;
}
}
return false;
}
int main()
{
int T;
scanf("%d",&T);
int flag=;
while(T--)
{
if(flag) puts("");
scanf("%d%d",&m,&n);
memset(vis,,sizeof vis);
vis[][]=;
a1[]=,a2[]=;
printf("Scenario #%d:\n",++flag);
if(!dfs(,,)) puts("impossible");
}
return ;
}