A Knight's Journey
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 40609 | Accepted: 13800 |
Description
Background
The knight (骑士) 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 (垂线) to this. The world of a knight is the chessboard (棋盘)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 (矩形的). Can you help this adventurous (爱冒险的) knight to make travel plans?
Problem
Find a path such that the knight (骑士) visits every square once. The knight can start and end on any square of the board.
Input
The
input
(投入) begins with a
positive
(积极的)
integer
(整数) 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
(棋盘), 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
(字母表): A, . . .
Output
The
output
(输出) for every
scenario
(方案) 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
(辞典编纂的) 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
(连结) 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
题意：
输入n,m. 即一个n*m的方格， 要求一个马（象棋里的马）能从任意地点开始，走完所有方格点。输出字典序最小的路径。
思路：
首先要明白，能从任意点遍历全图，那从第一个点也必须可以便利全图。因为要求字典序最小，所以我们就从第一个点DFS看看能否达成条件。
其次要注意，国际象棋横向是字母，竖向是数字。
第三，马走的顺序应该是x尽量小的前提下让y尽量小，才能保证字典序最小。
代码：
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
int p, q;
int path_x[27], path_y[27];
bool vis[27][27], flag;
int Move[][2] = {-1,-2, 1,-2, -2,-1, 2,-1, -2,1, 2,1, -1,2, 1,2};
void DFS(int x, int y, int step)
{
path_x[step] = x;
path_y[step] = y;
if(step == p*q)
{
flag = true;
return;
}
for(int i = 0;i < 8;i++)
{
int xx = x + Move[i][0];
int yy = y + Move[i][1];
if(xx<1||yy<1||xx>p||yy>q)
continue;
if(!vis[xx][yy] && !flag)
{
vis[xx][yy] = true;
DFS(xx,yy,step+1);
vis[xx][yy] = false;
}
}
}
int main()
{
int n;
cin>>n;
for(int T = 1; T <= n;T++)
{
memset(vis,false,sizeof(vis));
cin>>p>>q;
vis[1][1] = true;
flag = false;
DFS(1,1,1);
cout<<"Scenario #"<<T<<':'<<endl;
if(flag)
{
for(int i = 1;i <= p*q;i++)
printf("%c%d",path_y[i]+'A'-1,path_x[i]);
}
else
cout<<"impossible";
cout<<endl<<endl;
}
return 0;
}