http://poj.org/problem?id=2488
A Knight's Journey
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 24572 | Accepted: 8305 |
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
分析：
这里 1 <= p * q <= 26 题目数据量不是很大，可以直接暴力dfs求解，主要是保证输出第一个为最小字典序，这里我们控制遍历搜索时的顺序就可以了，这里主要体现在以下数组中：
int cy[]={-2,-2,-1,-1,1,1,2,2};
int cx[]={-1,1,-2,2,-2,2,-1,1};
这里cy数组的值为 {-2,-2,-1,-1,1,1,2,2} 因为cy为 输出串中的 A,B,C 故优先选择，然后cx {-1,1,-2,2,-2,2,-1,1} 一小一大的排列
这样就可以保证最小字典序
代码：
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 5 using namespace std; 6 7 //这样设置这两个数组很好的保证了最后输出的序列是字典序 8 int cy[]={-2,-2,-1,-1,1,1,2,2}; 9 int cx[]={-1,1,-2,2,-2,2,-1,1}; 10 11 int mark[30][30]; 12 int flag=0; 13 14 void dfs(int n,int m,int x,int y,int cnt) 15 { 16 int i,j,k; 17 if(flag) //序列找到便不用在找 18 { 19 return; 20 } 21 if(cnt==n*m) //找到序列进行输出 22 { 23 for(k=1;k<=n*m;k++) 24 for(i=1;i<=n;i++) 25 { 26 for(j=1;j<=m;j++) 27 { 28 if(k==mark[i][j]) 29 { 30 printf("%c%d",'A'-1+j,i); 31 } 32 } 33 } 34 printf("\n"); 35 flag=1; //标记表示序列 36 return; 37 } 38 for(i=0;i<8;i++) 39 { 40 int dx=x+cx[i]; 41 int dy=y+cy[i]; 42 if(dx>=1&&dx<=n&&dy>=1&&dy<=m&&!mark[dx][dy]) 43 { 44 mark[dx][dy]=cnt+1; //标记这点的位置状态 45 dfs(n,m,dx,dy,cnt+1); //进行下一个点的查找 46 mark[dx][dy]=0; //回溯 47 } 48 } 49 } 50 51 int main() 52 { 53 int t,k; 54 scanf("%d",&t); 55 for(k=1;k<=t;k++) 56 { 57 int n,m; 58 scanf("%d%d",&n,&m); 59 printf("Scenario #%d:\n",k); 60 if(n==1&&m==1) 61 { 62 printf("A1\n\n"); 63 continue; 64 }else if(n==1||m==1) 65 { 66 printf("impossible\n\n"); 67 continue; 68 } 69 memset(mark,0,sizeof(mark)); 70 int i,j; 71 for(i=1;i<=n;i++) 72 { 73 for(j=1;j<=m;j++) 74 { 75 mark[i][j]=1; 76 flag=0; 77 dfs(n,m,i,j,1); 78 if(flag) 79 { 80 break; 81 } 82 memset(mark,0,sizeof(mark)); 83 } 84 if(j<=m) 85 { 86 break; 87 } 88 } 89 if(i>n) 90 { 91 printf("impossible\n"); 92 } 93 printf("\n"); 94 } 95 return 0; 96 }