A Knight's Journey（搜索题）
A Knight's Journey
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Description
Background
The knight is getting bored of seeing the same black and white squares
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
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.
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.
he 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.
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：
按照骑士路线行走是否能将所有的点全部走完（一个小于8*8的棋盘）并按照字典序
来进行输出所走的路的坐标先输出列（A,B,C,D,E,F,G）
再输出行（1，2，3，4，5，6，7，8）；
由于要求按照字典许输出路线则按照
int dx[8] = {-1, 1, -2,2, -2, 2, -1, 1};
int dy[8] = {-2, -2, -1,-1, 1, 1, 2, 2};//为了使其按照字典序排列且先输出列则
按照Y从小到大走即可。（且X也要按照从小到大）
（在Y相同的情况下按照象限第三、第二、第四、第一的顺序来进行）
代码：
#include
#include
#include
#include
#include
using namespace std;
int map[88][88];
int vis[88][88];
int n,m;
int flag;
int dx[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
int dy[8] = {-2, -2, -1, -1, 1, 1, 2, 2};//为了使其按照字典序排列且 先输出列则按照Y从小到大走即可。（且X也要按照从小到大）（在Y相同的情况下按照象限第三、第二、第四、第一的顺序来进行）
void DFS(int x,int y,int step)
{
map[step][0]=x;
map[step][1]=y;
if(step==n*m)
{
flag=1;
return ;
}
for(int i=0; i<8; i++)
{
int nx=x+dx[i];
int ny=y+dy[i];
if(nx<1||nx>n||ny<1||ny>m||vis[nx][ny]||flag)
continue;
else
{
vis[nx][ny]=1;
DFS(nx,ny,step+1);
vis[nx][ny]=0;
}
}
}
int main()
{
int i,j;
int t;
cin>>t;
int countt=0;
while(t--)
{ flag=0;
scanf("%d%d",&n,&m);
memset(vis,0,sizeof(vis));
vis[1][1]=1;
DFS(1,1,1);
cout<<"Scenario #"<<++countt<<":"<