acm_Knight Moves

最新推荐文章于 2021-08-29 15:03:14 发布

清风徐来---

最新推荐文章于 2021-08-29 15:03:14 发布

阅读量260

点赞数

CC 4.0 BY-SA版权

分类专栏： ACM

本文链接：https://blog.youkuaiyun.com/xu_acm/article/details/51235971

ACM 专栏收录该内容

45 篇文章

订阅专栏

本文介绍了一种使用广度优先搜索算法解决国际象棋棋盘上骑士从一个位置到另一个位置的最短行走路径问题的方法。通过定义骑士的可能移动方式，并采用广度优先搜索策略，可以高效找到两方格间的最少移动次数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：

Problem Description

A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks that the most difficult part of the problem is determining the smallest number of knight moves between two given squares and that, once you have accomplished this, finding the tour would be easy. Of course you know that it is vice versa. So you offer him to write a program that solves the "difficult" part. Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest route from a to b.

Input

The input file will contain one or more test cases. Each test case consists of one line containing two squares separated by one space. A square is a string consisting of a letter (a-h) representing the column and a digit (1-8) representing the row on the chessboard.

Output

For each test case, print one line saying "To get from xx to yy takes n knight moves.".

Sample Input

e2 e4<br>a1 b2<br>b2 c3<br>a1 h8<br>a1 h7<br>h8 a1<br>b1 c3<br>f6 f6<br>

Sample Output

To get from e2 to e4 takes 2 knight moves.<br>To get from a1 to b2 takes 4 knight moves.<br>To get from b2 to c3 takes 2 knight moves.<br>To get from a1 to h8 takes 6 knight moves.<br>To get from a1 to h7 takes 5 knight moves.<br>To get from h8 to a1 takes 6 knight moves.<br>To get from b1 to c3 takes 1 knight moves.<br>To get from f6 to f6 takes 0 knight moves.<br>

题意：

就是一种走法，给你一种固定的走法，再给你一个起点坐标，一个终点坐标。。求最小步数。。

想法：

广搜，最先走到的就是最小的。。

代码：

#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
struct point
{
char x;
int y;
int moves;
};
queue<point> Q;
char e_x;
int e_y;
int map[305][305];
int dir[8][2]={{-2,-1},{-2,1},{-1,2},{1,2},{2,1},{2,-1},{1,-2},{-1,-2}};
int bfs(point s)
{
int i;
map[s.x][s.y]=1;
point head,t;
Q.push(s);
while(!Q.empty ())
{
head=Q.front ();
Q.pop ();
if(head.x==e_x&&head.y==e_y) return head.moves;
for(i=0;i<8;i++)
{
int x=head.x+dir[i][0];
int y=head.y+dir[i][1];
if(x>='a'&&x<='h'&&y>=1&&y<=8&&!map[x][y])
{
map[x][y]=1;
t.x=x;
t.y =y;
t.moves =head.moves+1;
Q.push(t);
}
}
}
}
int main()
{
int n,j,i,ans;
char s_x;
int s_y;
point start;
while(cin>>s_x>>s_y>>e_x>>e_y)
{
for(i='a';i<='h';i++)
for(j=1;j<=8;j++)
map[i][j]=0;
while(!Q.empty())
Q.pop ();
start.x=s_x;
start.y=s_y;
start.moves=0;
ans=bfs(start);
cout<<"To get from "<<s_x<<s_y<<" to "<<e_x<<e_y<<" takes "<<ans<<" knight moves."<<endl;
}
return 0;
}