UVa 439 - Knight Moves

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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 ));    }}