POJ 2488 A Knight's Journey
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A Knight's Journey
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 Source
TUD Programming Contest 2005, Darmstadt, Germany
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题目描述：
用大写字母表示行，数字表示列，判断马是否可以不重复的走完整个地图，输出字典序最小的序列，如果不可能输出impossible。
深度优先搜索考虑到字典序最小，所以直接从A1开始搜，但要按(-1,-2) (1,-2) (-2,-1) (2,-1) (-2,1) (2,1) (-1,2) (1,2)顺序方向进行搜索，才能保证字典序搜到的是最小的。
代码：
#include<stdio.h> #include<string.h> int road[80][2],vis[80][80]; //vis数组进行标记是否已经访问过 int n; int p,q; int dx[8] = {-1,1,-2,2,-2,2,-1,1},dy[8] = {-2,-2,-1,-1,1,1,2,2}; //int dx[8] = {-2,-2,-1,1,1,1,2,2},dy[8] = {-1,1,-2,2,-2,2,-1,1}; int flag; void dfs(int r,int c,int step) //当前 行，列，步数 { road[step][0] = r; road[step][1] = c; if(step == p * q) //如果能走完，步数应该为p*q { flag = 1; return ; } for(int i = 0;i < 8;i++) { int nx = r + dx[i],ny = c + dy[i]; if(nx >= 1 && nx <= p && ny >= 1 && ny <= q && vis[nx][ny] == 0 && !flag) //判断是否将要走的位置是否合法 { vis[nx][ny] = 1; dfs(nx,ny,step + 1); vis[nx][ny] = 0; } } } int main() { scanf("%d",&n); int k = 1; while(n--) { flag = 0; scanf("%d%d",&p,&q); memset(vis,0,sizeof(vis)); vis[1][1] = 1; dfs(1,1,1); printf("Scenario #%d:\n",k++); if(flag) for(int i = 1;i <= p * q;i++) { printf("%c%d",road[i][1] + 'A' - 1,road[i][0]); //进行一下转换 } else printf("impossible"); printf("\n"); if(n != 0) printf("\n"); } return 0; }