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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: 248864-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.
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
31 12 34 3
Sample Output
Scenario #1:A1Scenario #2:impossibleScenario #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 long14 #define pii pair<int,int>15 #define INF 0x3f3f3f3f16 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 }
POJ 2488 A Knight's Journey
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