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 3Sample Output
Scenario #1: A1 Scenario #2: impossible Scenario #3: A1B3C1A2B4C2A3B1C3A4B2C4Source
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/*大致题意：给出一个p行q列的国际棋盘，马可以从任意一个格子开始走，问马能否不重复的走完所有的棋盘。如果可以，输出按字典序排列最小的路径。打印路径时，列用大写字母表示（A表示第一列），行用阿拉伯数字表示（从1开始），先输出列，再输出行。
分析：如果马可以不重复的走完所有的棋盘，那么它一定可以走到A1这个格子。所以我们只需从A1这个格子开始搜索，就能保证字典序是小的；除了这个条件，我们还要控制好马每次移动的方向，控制方向时保证字典序最小（即按照下图中格子的序号搜索）。控制好这两个条件，直接从A1开始深搜就行了。
*/ Memory: 392K Time: 16MS Language: G++ Result: Accepted Source Code #include <string.h> #include <stdio.h> #include <algorithm> using namespace std; int vis[30][30]; int a[30][30]; int dir[8][2]={-1,-2,1,-2,-2,-1,2,-1,-2,1,2,1,-1,2,1,2}; int xx[30], yy[30]; int n, m; int flag; void init() { memset(vis, 0, sizeof(vis)); memset(xx, 0, sizeof(xx)); memset(yy, 0, sizeof(yy)); } void dfs ( int x, int y, int cnt ) { int i; if (cnt == n*m) { flag = 1; return ; } for ( i = 0;i < 8; i++ ) { int fx = x+dir[i][0]; int fy = y+dir[i][1]; if (fx > n||fy > m||fx <= 0||fy <= 0||vis[fx][fy] == 1||flag == 1) { continue; } else { vis[fx][fy] = 1; cnt++; xx[cnt] = fx; yy[cnt] = fy; dfs(fx, fy, cnt); vis[fx][fy] = 0; cnt--; } } } int main() { int i, j; int t; scanf ( "%d", &t ); for ( i = 1;i <= t; i++ ) { scanf ( "%d %d", &n, &m ); init(); flag = 0; vis[1][1] = 1; xx[1] = 1; yy[1] = 1; dfs(1, 1 ,1); printf ( "Scenario #%d:\n", i ); if ( flag == 1 ) { for ( j = 1;j <= n*m; j++ ) { printf ( "%c", yy[j]+'A'-1 ); printf ( "%d", xx[j] ); } printf ( "\n" ); } else printf ( "impossible\n" ); if (t != 0) printf ( "\n" ); } } 代码菜鸟，如有错误，请多包涵！！ 如果有帮助记得鼓励我一下，谢谢！！
