A - Knight Moves

HDU 1372

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 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.

这是一道BFS的经典题，注意国际象棋的骑士（马）的走法是走任意3*2格子的对角线，所以说有八个方向，可以用一个数组记录这八个方向；还有注意输出格式上的问题，’\n’不要忘记；

下面附关于BFS和queue的用法的博客：
BFS知识
http://blog.youkuaiyun.com/furturerock/article/details/5568305
C++ STL–queue 的使用方法
http://www.cnblogs.com/hdk1993/p/5809180.html

代码如下：

#include <algorithm>
#include <bitset>
#include <cassert>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#define INF 0x3f3f3f3f
#define EPS 1e-6
using namespace std;

int direction[8][2]={-1,2,-1,-2,1,2,1,-2,-2,1,-2,-1,2,1,2,-1};//8 directions
int chessboard[10][10],edx,edy;
int stx,sty;
typedef struct node{
    int x;
    int y;
    int step;
}LOC;//坐标
bool check(int x,int y){
    if(x<0||x>7||y<0||y>7||chessboard[x][y])return true;
    else return false;
}//判断是否结束
int BFS(){
    memset(chessboard,0,sizeof(chessboard));
    LOC p,q,next;
    p.x=stx;p.y=sty;p.step=0;
    chessboard[p.x][p.y]=1;
    queue<LOC>Q;
    Q.push(p);
    while(!Q.empty()){
        q=Q.front();
        Q.pop();
        if(q.x==edx && q.y==edy)return q.step;
        for(int i=0;i<8;i++){
            next.x=q.x+direction[i][0];
            next.y=q.y+direction[i][1];
            if(next.x==edx && next.y==edy)return q.step+1;//满足返回
            else if(check(next.x,next.y))continue;//越界返回
            else {chessboard[next.x][next.y]=1;
                next.step=q.step+1;
                Q.push(next);}//没有到头继续搜索
        }
    }
    return 0;
}
int main(){
    string a,b;
    while(cin>>a>>b){
        stx=a[0]-'a';
        sty=a[1]-'1';
        edx=b[0]-'a';
        edy=b[1]-'1';
       // printf("To get from %s to %s takes %d knight moves.\n", a, b,BFS());
        cout<<"To get from "<<a<<" to "<<b<<" takes "<<BFS()<<" knight moves."<<endl;

    }
    return 0;
}

//check()函数里面的判断很重要，要小心，x，y坐标要注意有没有超出棋盘边界，然后还要看之前搜索时有没有走过这个点。
回去困高，小桃肯定没有帮我打赏金。。好久不打天梯了。。。