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POJ 2488 A Knight's Journey

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A Knight‘s Journey

Time Limit: 1000ms
Memory Limit: 65536KB
This problem will be judged on PKU. Original ID: 2488
64-bit integer IO format: %lld      Java class name: Main
 
Background 
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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.
 

Sample Input

3
1 1
2 3
4 3

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

Source

 
 
解题:搜索。图中显示了移动方向,注意字典序小到大。
 
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 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 long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 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 }
View Code

 

POJ 2488 A Knight's Journey

标签:des   style   blog   http   color   os   io   java   ar   

原文地址:http://www.cnblogs.com/crackpotisback/p/3960871.html

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