A Knight's Journey

标签：adt   div   分析   oar   single   black   mst   pat   sam   

总时间限制: 1000ms 内存限制: 65536kB
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

大致题意：给出一个p行q列的国际棋盘，马可以从任意一个格子开始走，问马能否不重复的走完所有的棋盘。如果可以，输出按字典序排列最小的路径。打印路径时，列用大写字母表示（A表示第一列），行用阿拉伯数字表示（从1开始），先输出列，再输出行。

分析：如果马可以不重复的走完所有的棋盘，那么它一定可以走到A1这个格子。所以我们只需从A1这个格子开始搜索，就能保证字典序是小的；除了这个条件，我们还要控制好马每次移动的方向，控制方向时保证字典序最小（即按照下图中格子的序号搜索）。控制好这两个条件，直接从A1开始深搜就行了。

 1 #include<cstdio>  
 2 #include<cstring>  
 3 #include<algorithm>  
 4 using namespace std;  
 5   
 6 int path[88][88], vis[88][88], p, q, cnt;  
 7 bool flag;  
 8   
 9 int dx[8] = {-1, 1, -2, 2, -2, 2, -1, 1};  
10 int dy[8] = {-2, -2, -1, -1, 1, 1, 2, 2};  
11   
12 bool judge(int x, int y)  
13 {  
14     if(x >= 1 && x <= p && y >= 1 && y <= q && !vis[x][y] && !flag)  
15         return true;  
16     return false;  
17 }  
18   
19 void DFS(int r, int c, int step)  
20 {  
21     path[step][0] = r;  
22     path[step][1] = c;  
23     if(step == p * q)  
24     {  
25         flag = true;  
26         return ;  
27     }  
28     for(int i = 0; i < 8; i++)  
29     {  
30         int nx = r + dx[i];  
31         int ny = c + dy[i];  
32         if(judge(nx,ny))  
33         {  
34   
35             vis[nx][ny] = 1;  
36             DFS(nx,ny,step+1);  
37             vis[nx][ny] = 0;  
38         }  
39     }  
40 }  
41   
42 int main()  
43 {  
44     int i, j, n, cas = 0;  
45     scanf("%d",&n);  
46     while(n--)  
47     {  
48         flag = 0;  
49         scanf("%d%d",&p,&q);  
50         memset(vis,0,sizeof(vis));  
51         vis[1][1] = 1;  
52         DFS(1,1,1);  
53         printf("Scenario #%d:\n",++cas);  
54         if(flag)  
55         {  
56             for(i = 1; i <= p * q; i++)  
57                 printf("%c%d",path[i][1] - 1 + ‘A‘,path[i][0]);  
58         }  
59         else  
60             printf("impossible");  
61         printf("\n");  
62         if(n != 0)  
63             printf("\n");  
64     }  
65     return 0;  
66 }  

来源：http://blog.csdn.net/lyhvoyage/article/details/18355471

A Knight's Journey

标签：adt   div   分析   oar   single   black   mst   pat   sam   

原文地址：http://www.cnblogs.com/huashanqingzhu/p/7470997.html