2015 HUAS Summer Training#2~E

最新推荐文章于 2015-07-24 21:29:00 发布

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最新推荐文章于 2015-07-24 21:29:00 发布

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原文链接：http://www.cnblogs.com/chenchunhui/p/4671761.html

本文介绍了一个程序解决方案，用于计算国际象棋中马（骑士）从一个指定位置到另一个指定位置所需的最少移动次数。通过使用广度优先搜索算法（BFS），该程序能够高效地解决这个问题，为棋迷提供了一种简便的方法来探索骑士的走法。

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 Specification

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 Specification

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.

解题思路：该问题不需要求出该两点间的路径上的点，也不需要求出拥有该最短步数的路径个数，只需将其间的最短步数求出即可；该问题不仅需要求出两点间的最短路径，且要求出这样的最短路径有多少条和每一条路径上的每一点；找出从S出发到达D的所有路径长度小于6的路径，然后再从这些路径中寻找最短路径

程序代码：

#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;
}