[poj]2488 A Knight's Journey dfs+路径打印
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 45941 | Accepted: 15637 |
Description
![](http://poj.org/images/2488_1.jpg)
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
按照字典序输出路径,方向要按照字典序来搜索。
#include <iostream> #include <stdio.h> #include <cstring> #include <algorithm> using namespace std; bool v[10][10]; int p, q; int dir[8][2] = {{-2,-1},{-2,1},{-1,-2},{-1,2}, {1,-2},{1,2},{2,-1},{2,1}}; int px[100], py[100]; int step, flag; char R[8] = {'A','B','C','D','E','F','G','H'}; int dfs(int x, int y, int step) { if (step == p*q) { flag = 1; for (int i = 0; i < p*q; i++) { printf("%c%d", R[px[i]],py[i]+1); } printf("\n\n"); return 1; } int nx, ny; for (int i = 0; i < 8; i++) { nx = x + dir[i][0]; ny = y + dir[i][1]; if (!v[nx][ny] && nx>=0 && nx<q && ny>=0 && ny<p) { v[nx][ny] = 1; px[step] = nx; py[step] = ny; dfs(nx, ny, step+1); if (flag) return 1; //只搜索一次 v[nx][ny] = 0; } } return 0; } int main() { //freopen("1.txt", "r", stdin); //freopen("out.txt", "w", stdout); int T; int t = 0; cin >> T; while (T--) { cin >> p >> q; printf("Scenario #%d:\n", ++t); memset(v, 0, sizeof(v)); px[0] = 0; py[0] = 0; v[0][0] = 1; flag = 0; step = 0; if(!dfs(0, 0, 1)) printf("impossible\n\n"); } return 0; }