题目描述:
有一个 𝑛×𝑚n×m 的棋盘,在某个点 (𝑥,𝑦)(x,y) 上有一个马,要求你计算出马到达棋盘上任意一个点最少要走几步。
代码:
java
package lanqiao;
import java.util.*;
public class Main {
static int n,m,x,y;
static int[][] a = new int[410][410];
static int[] aa = new int[] {2, 1, 2, -1, -2, -1, -2, 1};
static int[] bb = new int[] {1, 2, -1, -2, -1, 2, 1, -2};
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
n = sc.nextInt();
m = sc.nextInt();
x = sc.nextInt();
y = sc.nextInt();
//初始化数组
for(int i = 1;i <= n;i ++)
{
for(int j = 1;j <= m;j ++)
{
a[i][j] = -1;
}
}
dfs(x,y,0);
a[x][y] = 0;
for(int i = 1;i <= n;i ++)
{
for(int j = 1;j <= m;j ++)
{
System.out.printf("%-5d", a[i][j]);
}
System.out.println();
}
}
public static void dfs(int x,int y,int t)
{
if(t >200) //DFS不加剪枝的话需要加阙值
{
return;
}
a[x][y] = t;
for(int i = 0;i < 8;i ++)
{
if(x + aa[i] >= 1 && y + bb[i] >= 1 && x + aa[i] <= n && y + bb[i] <= m
&& (a[x + aa[i]][y + bb[i]] == -1 || a[x + aa[i]][y + bb[i]] > t + 1))//需要对未走过的格子,或者新路线步数较短的格子进行重新赋值
{
dfs(x + aa[i],y + bb[i],t + 1);
}
}
}
}