HDU Knight Moves

最新推荐文章于 2025-08-12 21:03:53 发布

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最新推荐文章于 2025-08-12 21:03:53 发布

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文章标签： java 数据结构与算法

原文链接：http://www.cnblogs.com/newpanderking/archive/2011/09/19/2181705.html

本文介绍了一种使用广度优先搜索算法解决骑士在国际象棋盘上从一个位置到另一个位置所需的最少移动次数的问题。通过定义骑士可能的移动方向并采用队列进行逐层搜索，该算法能够有效地找到最短路径。

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Knight Moves

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 96 Accepted Submission(s): 77

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
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 f6

Sample Output

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

分析：广度优先搜索题

题意如图所示：一个棋子（骑士）可以有八个方向走，广搜确定最小的走的步数。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
int c[9][9];
int dir[8][2] = {{-2,-1},{-2,1},{-1,2},{1,2},{2,1},{2,-1},{1,-2},{-1,-2}};
typedef struct
{
int x,y,count;
}node;
node start,finish;
int bfs()
{
    memset(c,0,sizeof(c));
    node pre,cur;
    start.count = 0;
    queue<node> q;
    q.push(start);
    c[start.x][start.y] = 1;
while(!q.empty())
    {
        pre = q.front();
        q.pop();
if(pre.x == finish.x&&pre.y == finish.y)
return pre.count;
for(int i = 0; i < 8; i++)
        {
            cur.x = pre.x + dir[i][0];
            cur.y = pre.y + dir[i][1];
if(cur.x<1||cur.x>8||cur.y<1||cur.y>8)continue;
if(c[cur.x][cur.y]==1)continue;
            c[cur.x][cur.y] = 1;
            cur.count = pre.count + 1;
            q.push(cur);
        }
    }
return -1;
}
int main()
{
char row,end;
int col,ed;
int min;
while(scanf("%c",&row)!=EOF)
    {
        scanf("%d",&col);
        getchar();
        scanf("%c%d",&end,&ed);
        getchar();
        start.x = row-'a'+1;
        start.y = col;
        finish.x = end-'a'+1;
        finish.y = ed;
if(start.x==finish.x&&start.y==finish.y)
        min = 0;
else  min = bfs();
        printf("To get from %c%d to %c%d takes %d knight moves.\n",row,col,end,ed,min);
    }
return 0;
}