A Knight’s Journey
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标签：深度优先搜索
参考资料：相似题目：
题目
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.
输入
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, …
输出
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.
输入样例
3
1 1
2 3
4 3
输出样例
Scenario #1:
A1
Scenario #2:
impossible
Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4
解题思路
深度优先搜索，因为题目要求找到一条字典序最小的路径，所以搜索的方向要注意。
参考代码
#include<stdio.h>
#include<string.h>
#define MAXN 30
int board[MAXN][MAXN];
int dx[8]={-1, 1,-2, 2,-2, 2,-1, 1},
dy[8]={-2,-2,-1,-1, 1, 1, 2, 2};
int p,q;
int posx[MAXN],posy[MAXN];
int cnt;
int dfs(int x,int y){
board[x][y]=1;
posx[cnt]=x;
posy[cnt]=y;
cnt++;
if(cnt==p*q){
return 1;
}
for(int i=0;i<8;i++){
int nx=x+dx[i],ny=y+dy[i];
if(0<=nx && nx<p && 0<=ny && ny<q && board[nx][ny]==0){
if(dfs(nx,ny)==1)return 1;
board[nx][ny]=0;
cnt--;
}
}
return 0;
}
int main(){
int n;
scanf("%d",&n);
int i,j;
for(i=1;i<=n;i++){
memset(board,0,sizeof(board));
cnt=0;
scanf("%d%d",&p,&q);
int flag=dfs(0,0);
printf("Scenario #%d:\n",i);
if(flag==0){
printf("impossible\n");
}
else{
for(j=0;j<p*q;j++){
printf("%c%d",posy[j]+'A',posx[j]+1);
}
printf("\n");
}
printf("\n");
}
return 0;
}