POJ 2488 A Knight's Journey

A Knight's Journey

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on  PKU. Original ID: 2488
64-bit integer IO format: %lld      Java class name: Main

 

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

 

 

解题：搜索。图中显示了移动方向，注意字典序小到大。

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #include <stack>

13 #define LL long long

14 #define pii pair<int,int>

15 #define INF 0x3f3f3f3f

16 using namespace std;

17 const int dir[8][2] = {{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};

18 int n,m,path[100][2];

19 bool vis[100][100];

20 bool dfs(int x,int y,int cur){

21     if(cur >= n*m) return true;

22     for(int i = 0; i < 8; i++){

23         int tx = dir[i][0]+x;

24         int ty = dir[i][1]+y;

25         if(tx < 0 || tx >= n || ty < 0 || ty >= m || vis[tx][ty]) continue;

26         vis[tx][ty] = true;

27         path[cur][0] = tx+'A';

28         path[cur][1] = ty+1;

29         if(dfs(tx,ty,cur+1)) return true;

30         vis[tx][ty] = false;

31     }

32     return false;

33 }

34 int main() {

35     int t,k = 1;

36     scanf("%d",&t);

37     while(t--){

38         scanf("%d %d",&m,&n);

39         bool flag = false;

40         memset(vis,false,sizeof(vis));

41         path[0][0] = 'A';

42         path[0][1] = 1;

43         vis[0][0] = true;

44         printf("Scenario #%d:\n",k++);

45         if(dfs(0,0,1)){

46             for(int i = 0; i < n*m; i++) printf("%c%d",path[i][0],path[i][1]);

47             puts("");

48         }else puts("impossible");

49         puts("");

50     }

51     return 0;

52 }

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