18.06.30 POJ 2488:A Knight's Journey

描述

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.

输入

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, . ..

输出

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.

样例输入

3
1 1
2 3
4 3

样例输出

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

来源

TUD Programming Contest 2005, Darmstadt, Germany

 1 #include <cstdio>
 2 #include <string>
 3 #include <memory.h>
 4 #include <algorithm>
 5 #include <stdlib.h>
 6 #include <math.h>
 7 #include <iostream>
 8 #include<queue>
 9 #include <vector>
10 #include <bitset>
11 using namespace std;
12 
13 int r, c;
14 int kase;
15 int visited[10][10];
16 int jing = 0;
17 bool flag;
18 char sign[9] = { 0,'A','B','C','D','E','F','G','H' };
19 int dir1[8] = { -2,-2,-1,-1,1,1,2,2 }, dir2[8] = {-1,1,-2,2,-2,2,-1,1};
20 struct node {
21     int x, y;
22     node(int a, int b) :x(a), y(b) {}
23 };
24 vector<node> solution;
25 void dfs(int x,int y ) {
26     if (jing == r * c) {
27         printf("Scenario #%d:\n", kase);
28         vector<node>::iterator i1 = solution.begin(), i2 = solution.end();
29         for (; i1 != i2; i1++) 
30             cout << sign[(*i1).x] << (*i1).y;
31         printf("\n\n");
32         flag = true;
33         return;
34     }
35     if (flag)return;
36     for (int i = 0; i <= 7; i++) {
37         int xx = x + dir1[i], yy = y + dir2[i];
38         if (visited[xx][yy]||xx<1||xx>r||yy<1||yy>c)
39             continue;
40         visited[xx][yy] = 1;
41         solution.push_back(node(xx, yy));
42         jing++;
43         dfs(xx, yy);
44         if (flag == true)
45             return;
46         solution.pop_back();
47         visited[xx][yy] = 0;
48         jing--;
49     }
50 }
51 
52 int main()
53 {
54     int t;
55     scanf("%d", &t);
56     for(kase=1;kase<=t;kase++){
57         scanf("%d%d", &c, &r);
58         solution.clear();
59         memset(visited, 0, sizeof(int) * 10 * 10);
60         jing = 1;
61         flag = false;
62         visited[1][1] = 1;
63         solution.push_back(node(1, 1));
64         dfs(1, 1);
65         if (flag == false) {
66             printf("Scenario #%d:\n", kase);
67             printf("impossible\n\n");
68         }
69     }
70     return 0;
71 }

View Code

输出格式