POJ 2488 A Knight’s Journey C语言版

POJ题解 4695浏览

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

Slyar:说一下题目大意。给出一个国际棋盘的大小，判断马能否不重复的走过所有格，并记录下其中按字典序排列的第一种路径。

“马的遍历”是一道经典回溯题，当然还是DFS…这题有3个要密切注意的地方：

1、题目要求以”lexicographically”方式输出，也就是字典序…一开始没看懂这个词结果WA了N次…要以字典序输出路径，那么方向数组就要以特殊的顺序排列了…这样只要每次从dfs(1,1)开始搜索，第一个成功遍历的路径一定是以字典序排列…

2、国际象棋的棋盘，横行为字母，表示横行坐标的是y；纵行为数字，表示纵行的坐标是x…一开始又搞反了…

3、虽然题目没说，但是这道题最后一组数据后是没有空行的…否则会PE…

```#include
#include

#define MAX 27

int map[MAX][MAX], sx[MAX], sy[MAX];
/* 字典序坐标数组 */
int dir[8][2] = {{-2, -1}, {-2, 1}, {-1, -2}, {-1, 2},{1, -2}, {1, 2}, {2, -1}, {2, 1}};
int p, q, sign, step;

void dfs(int i, int j)
{
/* 如果已经遍历完成则直接返回 */
if (sign) return;
int x, y, k;
/* 步数 + 1 */
step++;
/* 记录当前坐标 */
sx[step] = i;
sy[step] = j;
/* 当步数等于棋盘格子数则成功 */
if(step == p * q)
{
sign = 1;
return;
}
/* 置已访问标记 */
map[i][j] = 1;
for (k = 0; k < 8; k++)
{
/* 第一个坐标是y方向的，第二个坐标是x方向的 */
y = j + dir[k][0];
x = i + dir[k][1];
/* 格子未被访问且不越界则向下搜索 */
if (map[x][y] == 0 && x > 0 && x <= p && y > 0 && y <= q)
{
dfs(x, y);
/* 不符合条件回溯时步数 - 1 */
step--;
}
}
/* 返回时重置未访问标记 */
map[i][j] = 0;
}

int main()
{
int i, j, n, t = 0;
scanf("%d", &n);
while(n--)
{
sign = 0;
step = 0;
t++;
scanf("%d%d", &p, &q);
for (i = 1; i <= p; i++)
{
for(j = 1; j <= q; j++)
{
map[i][j] = 0;
}
}
dfs(1, 1);
printf("Scenario #%d:\n", t);
if (sign)
{
for (i = 1; i <= p * q; i++)
{
/* 注意棋盘行(y)是字母，列(x)是数字 */
printf("%c%d", sy[i] + 64, sx[i]);
}
printf("\n");
}
else
{
printf("impossible\n");
}
/* 最后一组数据没空行 */
if (n != 0) printf("\n");
}
//system("pause");
return 0;
}
```

网友最新评论 (9)

1. 这个递归里 sign 不多余么 ？ 不是在 if(step == p * q) { sign = 1; return; } 这句就已经结束了 ？
sound9年前 (2011-09-26)回复
2. 呵呵，写的太好了，每每看到都大彻大悟，非常感谢你……
fenghuijun10年前 (2011-05-13)回复
3. @gongzhi 我继续不解释，自己考虑...
Slyar11年前 (2010-05-27)回复
4. /* 当步数等于棋盘格子数则成功 */ if(step == p * q) { sign = 1; return; } 这里步数其实每次应该每次加3吧 还有就是字典序应该还有一个吧 应该有8个走法啊
gongzhi11年前 (2010-05-27)回复
5. 谢谢 我知道了
zeot11年前 (2010-05-19)回复
6. @zeot Y也不是随便，数字大小也属于字典序...
Slyar11年前 (2010-05-18)回复
7. 哦 是不是x的变化顺序是-2 -1 1 2 而y随便
zeot11年前 (2010-05-18)回复
8. @zeot 字典序啊，你知道英文字典的排序都是从A-Z的吧，这里也一样，输出的时候要先输出A，然后是B、C、D...只有按那个方向数组来搜，才会自动按照A-Z输出，你想想...
Slyar11年前 (2010-05-18)回复
9. 请问下 字典序{-2, -1}, {-2, 1}, {-1, -2}, {-1, 2},{1, -2}, {1, 2}, {2, -1}和这个有什么关系 字典序是什么意思 谢谢
zeot11年前 (2010-05-18)回复