POJ 2488 A Knight's Journey【DFS + 回溯应用】

我的链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=19651#problem/A

A Knight's Journey

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23452		Accepted: 7944

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

算法：DFS + 回溯

题意：骑士周游列国问题，说白了就是任一给你一个棋盘（当然满足格子数小于26）

让你从第一个格子按照图中所给的马走日的步法

不重复遍历走完整个棋盘。然后按照字典序输出路径即可

思路：按照图中的八个方向dfs即可。

注意: 1.回溯的用法，千万别忘了还原状态。

2.路径的输出，字典序的是列。。。

//Accepted	172 KB	16 ms	C++	1191 B	2013-03-02 21:12:01
#include<cstdio>
#include<cstring>

const int maxn = 30;

int w[maxn*maxn][2]; //存储路径 
bool vis[maxn][maxn];
int dir[8][2] = {-2,-1, -2,1, -1,-2, -1,2, 1,-2, 1,2, 2,-1, 2,1};

int p,q;
int n;
int sum;

bool search(int row, int col)
{
	sum++;
	w[sum][0] = row; w[sum][1] = col;
	if(sum == n) return true;
	
	else for(int i = 0; i < 8; i++)
	{
		int next_r = row + dir[i][0];
		int next_c = col + dir[i][1];
		
		if(next_r >= 1 && next_r <= q && next_c >= 1 && next_c <= p && !vis[next_r][next_c])
		{ //注意行和列的范围 
			vis[next_r][next_c] = true;  
			if(search(next_r, next_c)) return true; //
			vis[next_r][next_c] = false; 
			sum--;  //注意点 开始WA了好久 
		} 
	}
	return false;
}
int main()
{
	int T;
	scanf("%d", &T);
	for(int test = 1; test <= T; test++)
	{
		bool result = false;
		scanf("%d%d", &p, &q);
		
		n = p*q; sum = 0;
		if(n >= 1 && n <= 26)
		{
			memset(w,0,sizeof(w));
			memset(vis,false,sizeof(vis));
			
			vis[1][1] = true;
			 
			result = search(1,1);
			
			printf("Scenario #%d:\n", test);
			if(result)
			{
				for(int i = 1; i <= n; i++)
					printf("%c%d", w[i][0]+'A'-1, w[i][1]);
				printf("\n");
			}
			else
			{
				printf("impossible\n");
			}
			 
			if(test != T) printf("\n"); //注意格式 
		}
	}
	return 0;
}