Description
Background The knight (骑士) 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 (垂线) to this. The world of a knight is the chessboard (棋盘)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 (矩形的). Can you help this adventurous (爱冒险的) knight to make travel plans? Problem Find a path such that the knight (骑士) visits every square once. The knight can start and end on any square of the board.Input
The input (投入) begins with a positive (积极的) integer (整数) 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 (棋盘), 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 (字母表): A, . . .Output
The output (输出) for every scenario (方案) 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 (辞典编纂的) 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 (连结) 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: A1B3C1A2B4C2A3B1C3A4B2C4题意：
输入n,m. 即一个n*m的方格， 要求一个马（象棋里的马）能从任意地点开始，走完所有方格点。输出字典序最小的路径。
思路：
首先要明白，能从任意点遍历全图，那从第一个点也必须可以便利全图。因为要求字典序最小，所以我们就从第一个点DFS看看能否达成条件。
其次要注意，国际象棋横向是字母，竖向是数字。
第三，马走的顺序应该是x尽量小的前提下让y尽量小，才能保证字典序最小。
代码：
#include <iostream> #include <string.h> #include <stdio.h> using namespace std; int p, q; int path_x[27], path_y[27]; bool vis[27][27], flag; int Move[][2] = {-1,-2, 1,-2, -2,-1, 2,-1, -2,1, 2,1, -1,2, 1,2}; void DFS(int x, int y, int step) { path_x[step] = x; path_y[step] = y; if(step == p*q) { flag = true; return; } for(int i = 0;i < 8;i++) { int xx = x + Move[i][0]; int yy = y + Move[i][1]; if(xx<1||yy<1||xx>p||yy>q) continue; if(!vis[xx][yy] && !flag) { vis[xx][yy] = true; DFS(xx,yy,step+1); vis[xx][yy] = false; } } } int main() { int n; cin>>n; for(int T = 1; T <= n;T++) { memset(vis,false,sizeof(vis)); cin>>p>>q; vis[1][1] = true; flag = false; DFS(1,1,1); cout<<"Scenario #"<<T<<':'<<endl; if(flag) { for(int i = 1;i <= p*q;i++) printf("%c%d",path_y[i]+'A'-1,path_x[i]); } else cout<<"impossible"; cout<<endl<<endl; } return 0; }