POJ - 2488：A Knight's Journey

A Knight’s Journey

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标签：深度优先搜索
参考资料：

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题目

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.

输入

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, . . .

输出

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.

输入样例

3
1 1
2 3
4 3

输出样例

Scenario #1:
A1


Scenario #2:
impossible


Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

解题思路

深度优先搜索，因为题目要求找到一条字典序最小的路径，所以搜索的方向要注意。

参考代码

#include
#include
#define MAXN 30
int board[MAXN][MAXN];
int dx[8]={-1, 1,-2, 2,-2, 2,-1, 1},
    dy[8]={-2,-2,-1,-1, 1, 1, 2, 2};
int p,q;
int posx[MAXN],posy[MAXN];
int cnt;
int dfs(int x,int y){
	board[x][y]=1;
	posx[cnt]=x;
	posy[cnt]=y;
	cnt++;
	if(cnt==p*q){
		return 1;
	}
	for(int i=0;i<8;i++){
		int nx=x+dx[i],ny=y+dy[i];
		if(0<=nx && nx<p && 0<=ny && ny<q && board[nx][ny]==0){
			if(dfs(nx,ny)==1)return 1;
			board[nx][ny]=0;
			cnt--;
		}
	}
	return 0;
}
int main(){
	int n;
	scanf("%d",&n);
	int i,j;
	for(i=1;i<=n;i++){
		memset(board,0,sizeof(board));
		cnt=0;
		scanf("%d%d",&p,&q);
		int flag=dfs(0,0);
		printf("Scenario #%d:\n",i);
		if(flag==0){
			printf("impossible\n");
		}
		else{
			for(j=0;j<p*q;j++){
				printf("%c%d",posy[j]+'A',posx[j]+1);
			}
			printf("\n");
		}
		printf("\n");	
	}
	return 0;
}