文章作者：姜南(Slyar) 文章来源：Slyar Home (www.slyar.com) 转载请注明，谢谢合作。
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...
其他内容注释里很清楚了...优化了代码，0MS...
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#include <stdio.h> #include <stdlib.h> #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; } |
