UVa 439 - Knight Moves

题目：给出国际象棋棋盘中的两个点，求马从一个点跳到另一个点的最少步数。

分析：简单题、搜索、最短路。利用bfs去接即可。

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>

using namespace std;

typedef struct qnode 
{
    int x,y,d;
    qnode( int a, int b, int c ){x = a;y = b;d = c;}
    qnode(){}
}queue;
queue Q[90];

int  maps[9][9];
int  d[8][2]={{2,1},{2,-1},{-1,2},{1,2},{-2,1},{-2,-1},{-1,-2},{1,-2}};

int bfs( int x , int y, int xx, int yy )
{
    if  ( x == xx && y == yy ) return 0;
    memset( maps, 0, sizeof(maps) );
    
    int move = 0,save = 1;
    Q[0] = queue( x, y, 0 );
    while ( move < save ) {
		queue now = Q[move ++];                 
        for ( int k = 0 ; k < 8 ; ++ k ) {
            x = now.x + d[k][0];
            y = now.y + d[k][1];
            if ( !maps[x][y] && x > 0 && x < 9 && y > 0 && y < 9 ) {
                if ( x == xx && y == yy  )
                   return now.d+1;
                Q[save ++] = queue( x, y, now.d+1 );
                maps[x][y] = 1;
            }
        }
	}
	return -1;
}

int main()
{
    char ch1,ch2;
    int  x1,x2,y1,y2;
    while ( cin >> ch1 >> y1 >> ch2 >> y2 ) {
        x1 = (int)(ch1-'a'+1);
        x2 = (int)(ch2-'a'+1);
        printf("To get from %c%d to %c%d takes %d knight moves.\n",ch1,y1,ch2,y2,bfs( x1 , y1, x2, y2 ));
    }
}