A Knight's Journey POJ - 2488

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?

Find a path such that the knight visits every square once. The knight can start and end on any square of the board.

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, . . .

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.

3
1 1
2 3
4 3

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

一个骑士要走完棋盘上所有的点一次且仅一次,起点和终点不限,求走的顺序并输出对应的下标,而且尽可能的优先输出字典序最小的那个.

典型的搜索,只是每次搜索的顺序要使字典序最小的顺序来搜索,然后用数组记录路径就OK了.

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<iostream>
using namespace std;

int n,m;
int path[67];
char pat[67];
bool vis[65][65];
bool check;
int dx[]={-2,-2,-<
4000
span class="hljs-number">1,-1,+1,+1,2,2}; // 骑士走的位置按字典序最小的来排列
int dy[]={-1,1,-2,+2,-2,+2,-1,1};

void dfs(int x,int y,int k)
{
if(k==m*n)
check=true;
if(check)
return ;
for(int i=0;i<8;++i){
int sx=x+dx[i];
int sy=y+dy[i];
if(sx>0&&sy>0&&sx<=n&&sy<=m&&vis[sx][sy]==false){
vis[sx][sy]=true;
path[k+1]=sy;
pat[k+1]='A'+sx-1;
dfs(sx,sy,k+1);
vis[sx][sy]=false;
if(check)
return ;
}
}
}

int main()
{
int cas=0;
int t;
scanf("%d",&t);
while(t--){
scanf("%d %d",&n,&m);
swap(n,m);
memset(vis,false,sizeof(vis));
check=false;
for(int i=1;i<=n;++i){
for(int j=1;j<=m;++j){
path[1]=j;
pat[1]='A'+i-1;
vis[i][j]=true;
dfs(i,j,1);//从某一点开始搜索
vis[i][j]=false;
if(check)//如果搜到就输出解
break;
}
if(check)
break;
}
cout<<"Scenario #"<<++cas<<':'<<endl;
if(check)
for(int i=1;i<=n*m;++i)
cout<<pat[i]<<path[i];
else cout<<"impossible";
cout<<endl;
if(t)
cout<<endl;
}
return 0;
}