HDU 1372（只是复习一下BFS）

Knight Moves

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

Total Submission(s): 7948 Accepted Submission(s): 4688



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.

Source

University of Ulm Local Contest 1996



题意：现有一个8*8棋盘，给你一个起始坐标和终点坐标，问你一个“马”（中国象棋）从起始坐标到终点坐标至少需要走几步。

code:

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
#define n 8
#define m 8
int t_x, t_y, vis[10][10];
int d_x[] = {-1,-2,-2,-1,1,2,2,1}; //"马"的步伐，摩擦摩擦，似魔鬼的步伐
int d_y[] = {-2,-1,1,2,-2,-1,1,2};
struct Node
{
    int x, y, step;
}node;

int bfs(int x, int y)
{
    memset(vis, 0, sizeof(vis));
    queue<Node> q;
    node.x = x, node.y = y, node.step = 0;
    q.push(node); vis[x][y] = 1;
    while(!q.empty())
    {
        Node u = q.front(); q.pop();
        if(u.x == t_x && u.y == t_y) return u.step;
        for(int i=0; i<8; i++)
        {
            Node next;
            next.x = u.x+d_x[i], next.y = u.y+d_y[i];
            if(vis[next.x][next.y] || next.x < 1 || next.x > n || next.y < 1 || next.y > m) continue;
            next.step = u.step+1;
            q.push(next); vis[next.x][next.y] = 1;
        }
    }
}

int main()
{
    char s[5], t[5];

    while(~scanf("%s%s", s,t))
    {
        int s_x = s[1]-'0';
        int s_y = s[0]-'a'+1;
        t_x = t[1]-'0';
        t_y = t[0]-'a'+1;
        printf("To get from %s to %s takes %d knight moves.\n", s,t,bfs(s_x,s_y));
    }
    return 0;
}