A Knight‘s Journey
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 30656 | Accepted: 10498 |
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.
Sample Input
3 1 1 2 3 4 3
Sample Output
Scenario #1: A1 Scenario #2: impossible Scenario #3: A1B3C1A2B4C2A3B1C3A4B2C4
Source
TUD Programming Contest 2005, Darmstadt, Germany
题目思路很清晰就是说如果能遍历整个图则打印路径，如果不行，则输出impossible
这道题按照字典序去搜索，想清楚行列关系再写，不然会WA
1 #include<cstdio>
2 #include<cstring>
3 #include<stdlib.h>
4 #include<algorithm>
5 using namespace std;
6 const int MAXN=200+10;
7 const int INF=0x3f3f3f3f;
8 int n,m,flag;
9 int vis[MAXN][MAXN],vx[MAXN],vy[MAXN];
10 int dir[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};//字典序
11 void DFS(int x,int y,int sum)
12 {
13 vx[sum]=x;//列，字母
14 vy[sum]=y;//行，数字
15 if(sum==n*m)
16 {
17 flag=1;
18 return ;
19 }
20 for(int i=0;i<8;i++)
21 {
22 int xx=x+dir[i][0];
23 int yy=y+dir[i][1];
24 if((1<=xx&&xx<=m)&&(1<=yy&&yy<=n)&&!vis[xx][yy]&&!flag)
25 {
26 vis[xx][yy]=1;
27 DFS(xx,yy,sum+1);
28 vis[xx][yy]=0;
29 }
30 }
31 }
32 int main()
33 {
34 //freopen("in.txt","r",stdin);
35 int kase,cnt=0;
36 scanf("%d",&kase);
37 while(kase--)
38 {
39 flag=0;
40 scanf("%d %d",&n,&m);
41 memset(vis,0,sizeof(vis));
42 memset(vx,0,sizeof(vx));
43 memset(vy,0,sizeof(vy));
44 vis[1][1]=1;
45 DFS(1,1,1);
46 printf("Scenario #%d:\n",++cnt);
47 if(flag)
48 {
49 for(int i=1;i<=n*m;i++)
50 printf("%c%d",‘A‘+vx[i]-1,vy[i]);
51 printf("\n");
52 }
53 else
54 printf("impossible\n");
55 printf("\n");
56
57 }
58 return 0;
59 }
POJ 2488 A Knight's Journey (DFS),布布扣,bubuko.com
POJ 2488 A Knight's Journey (DFS)