poj 2488 A Knight's Journey
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A Knight's Journey
Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 28501 Accepted: 9745
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
31 12 34 3
Sample Output
Scenario #1:A1Scenario #2:impossibleScenario #3:A1B3C1A2B4C2A3B1C3A4B2C4
Source
TUD Programming Contest 2005, Darmstadt, Germany
好久没做dfs，再加上突然搞出来一个字典序，把爹做出翔
其实大家不要被什么字典序唬住了，字典序方面只要处理一些细节就好，主要的还是dfs，相当简单的dfs，全题个人认为有两个比较值得学习的地方，详见代码：
#include<algorithm>#include<iostream>#include<cstdio>#include<cstring>using namespace std;int dir[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};///本题第一个撸点，因为题目要求要按照字典序输出，所以要先搜索字典序较小的点，即在使坐标第一个值尽量小的情况下再使第二个坐标尽量小。int p, q;bool used[100][100];int path[100][100];bool flag;bool check(int x, int y){if(x > 0 && y > 0 && x <= p && y <= q && !used[x][y] && !flag)return 1;return 0;}void dfs(int x, int y, int step){path[step][0] = x;path[step][1] = y;if(step == p*q){///printf("huang\n");flag = 1;return;}for(int i=0; i<8; ++i){int xx = x + dir[i][1];int yy = y + dir[i][0];if(check(xx , yy)){used[xx][yy] = 1;dfs(xx, yy, step+1);used[xx][yy] = 0;}}}int main(){int cnt, T;scanf("%d", &T);cnt = 0;while(T--){cnt++;flag = 0;scanf("%d%d", &p, &q);memset(used, 0, sizeof(used));used[1][1]=1;dfs(1, 1, 1);/**本题第二个撸点，因为如果棋子能够走遍整个棋盘，那么一定能走到A1，那么根据从字典序较小的点开始搜索的规律，所有情况都应该从A1开始搜索**/printf("Scenario #%d:\n", cnt);if(!flag) printf("impossible\n");else if(flag){for(int i=1; i<=p*q; ++i)printf("%c%d", path[i][1]- 1 + 'A', path[i][0]);printf("\n");}if(T != 0) printf("\n");}return 0;}0 0
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