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.
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
题目链接：http://poj.org/problem?id=2488
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 using namespace std; 5 int dir[][2]={{-1,-2},{1,-2},{-2,-1},{2,-1},{-2,1},{2,1},{-1,2},{1,2}}; 6 bool flag[27][27]; 7 struct node 8 { 9 int x,y; 10 }dd[27]; 11 12 int p,q; 13 int maxSize; 14 bool mark; 15 void init() 16 { 17 memset(flag,false,sizeof(flag)); 18 maxSize=p*q; 19 mark=false; 20 } 21 bool judge(int x,int y) 22 { 23 if(x<1 || x>p || y<1 || y>q || flag[x][y]) 24 { 25 return false; 26 } 27 return true; 28 } 29 void dfs(int x,int y,int step) 30 { 31 if(step==maxSize) 32 { 33 int i; 34 for(i=0;i<step;i++) 35 { 36 printf("%c%d",'A'+dd[i].y-1,dd[i].x); 37 } 38 printf("\n"); 39 mark=true; 40 return; 41 } 42 if(mark) 43 return; 44 int i; 45 for(i=0;i<8;i++) 46 { 47 int xx=x+dir[i][0]; 48 int yy=y+dir[i][1]; 49 if(!judge(xx,yy)) 50 { 51 continue; 52 } 53 dd[step].x=xx; 54 dd[step].y=yy; 55 flag[xx][yy]=true; 56 dfs(xx,yy,step+1); 57 if(mark) 58 return; 59 dd[step].x=0; 60 dd[step].y=0; 61 flag[xx][yy]=false; 62 } 63 } 64 int main() 65 { 66 int t; 67 scanf("%d",&t); 68 int cnt=1; 69 while(t--) 70 { 71 scanf("%d%d",&p,&q); 72 init(); 73 int i,j; 74 printf("Scenario #%d:\n",cnt++); 75 for(i=1;i<=p;i++) 76 { 77 for(j=1;j<=q;j++) 78 { 79 flag[i][j]=true; 80 dd[0].x=i; 81 dd[0].y=j; 82 dfs(i,j,1); 83 flag[i][j]=false; 84 dd[0].x=0; 85 dd[0].y=0; 86 if(mark) 87 break; 88 } 89 if(mark) 90 break; 91 } 92 if(!mark) 93 { 94 printf("impossible\n"); 95 } 96 printf("\n"); 97 } 98 return 0; 99 }
此题简单DFS，不过，还是花了不少时间，题目要求输出的路径（在此为一个字符串）是字典序最小，因此，dir[][2]方向数组的顺序就必须严格要求，其他的没什么陷阱