POJ-2488 A Knight's Journey(需注意搜索顺序的深搜)

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 26950		Accepted: 9195

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

Source

TUD Programming Contest 2005, Darmstadt, Germany

思路:

         深度优先搜索,但因为最后要按字典序输出走过的路径,所以需要注意搜索方向的顺序问题；

         

代码:

#include <stdio.h>
#include <string.h>
#include <stack>
#define N 30

using namespace std;

struct Node{
	int x;
	int y;
};

int m, n, count;
char map[N][N];
int fx[8][2] = {{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};
stack<Node>q;

int dfs(Node cur);

int main()
{
	int k, t = 0;
	scanf("%d", &k);
	while(k --){
		t ++;
		while(!q.empty()) q.pop();
		scanf("%d%d", &n, &m);
		memset(map, '-', sizeof(map));
		for(int i = 1; i <= m; i ++){
			for(int j = 1; j <= n; j ++){
				map[i][j] = '0';
			}
		}
		count = 1;
		Node start;
		start.x = 1;
		start.y = 1;
		map[1][1] = '-';
		int ok = dfs(start);
		printf("Scenario #%d:\n", t);
		if(!ok) printf("impossible");
		else{
			q.push(start);
			while(!q.empty()){
				printf("%c%d", q.top().x + 'A' - 1, q.top().y);
				q.pop();
			}
		}
		printf("\n\n");
	}

	return 0;
}

int dfs(Node cur)
{
	if(count == m * n) return 1;
	for(int i = 0; i < 8; i ++){
		Node nw = cur;
		nw.x += fx[i][0];
		nw.y += fx[i][1];
		if(nw.x <= 0 || nw.y <= 0 || map[nw.x][nw.y] == '-') continue;
		else{
			count ++;
			map[nw.x][nw.y] = '-';
			int ok = dfs(nw);
			if(ok == 1){
				q.push(nw);
			   	return 1;
			}
			count --;
			map[nw.x][nw.y] = '0';
		}
	}

	return 0;
}