POJ 1915 Knight Moves【广搜】

最新推荐文章于 2022-05-12 00:26:09 发布

weixin_33971205

最新推荐文章于 2022-05-12 00:26:09 发布

阅读量476

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/dream-wind/archive/2012/03/15/2397832.html

本文探讨了如何使用程序解决骑士在棋盘上从起点到终点的最短路径问题，通过分析给定的示例输入和输出，阐述了解决方案并提供了关键代码实现，帮助读者理解算法逻辑和优化思路。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description

Background
Mr Somurolov, fabulous chess-gamer indeed, asserts that no one else but him can move knights from one position to another so fast. Can you beat him?
The Problem
Your task is to write a program to calculate the minimum number of moves needed for a knight to reach one point from another, so that you have the chance to be faster than Somurolov.
For people not familiar with chess, the possible knight moves are shown in Figure 1.

Input

The input begins with the number n of scenarios on a single line by itself.
Next follow n scenarios. Each scenario consists of three lines containing integer numbers. The first line specifies the length l of a side of the chess board (4 <= l <= 300). The entire board has size l * l. The second and third line contain pair of integers {0, ..., l-1}*{0, ..., l-1} specifying the starting and ending position of the knight on the board. The integers are separated by a single blank. You can assume that the positions are valid positions on the chess board of that scenario.

Output

For each scenario of the input you have to calculate the minimal amount of knight moves which are necessary to move from the starting point to the ending point. If starting point and ending point are equal,distance is zero. The distance must be written on a single line.

Sample Input

3
8
0 0
7 0
100
0 0
30 50
10
1 1
1 1

Sample Output

5
28

code：

View Code

#include<stdio.h>
#include<string.h>
int s[301][301];
int vis[301][301];
int step[301][301];
int f[16]={-1,-2,-2,-1, 1,2,2,1, -2,-1,1,2, 2,1,-1,-2};
struct node
{
int x;
int y;
}q[100000];
int main()
{
int front,rear,t,i,j,n,x0,y0,x1,y1;
    scanf("%d",&t);
while(t--)
    {
        memset(vis,0,sizeof(vis));
        memset(step,0,sizeof(step));
        scanf("%d%d%d%d%d",&n,&x0,&y0,&x1,&y1);
        front=0;
        rear=0;
        q[rear].x=x0;
        q[rear++].y=y0;
        step[x0][y0]=0;
        vis[x0][y0]=1;
while(front<rear)
        {
for(i=0;i<8;i++)
if(!vis[q[front].x+f[i]][q[front].y+f[i+8]]&&q[front].x+f[i]>=0&&q[front].x+f[i]<n&&q[front].y+f[i+8]>=0&&q[front].y+f[i+8]<n)    
                {
                    q[rear].x=q[front].x+f[i];
                    q[rear++].y=q[front].y+f[i+8];
                    vis[q[front].x+f[i]][q[front].y+f[i+8]]=1;
                    step[q[front].x+f[i]][q[front].y+f[i+8]]=step[q[front].x][q[front].y]+1;
if(vis[x1][y1])
break;
                }
            front++;
if(vis[x1][y1])
break;
        }
        printf("%d\n",step[x1][y1]);
    }
return 0;
}