北大 poj 2243 Knight moves

这是我第一次做广搜的题，呃，错了无数次，原因有二：一是因为对广搜的形式理解不好，不知道怎么去记忆路径，第二是忘了原来骑士走的是“日”字形……也就是先向某个方向走2步，再向左或右走一步。所以一共有八个方向……

 

/* THE ROGRAM IS MADE BY PYY */ /*----------------------------------------------------------------------------// URL : http://poj.org/problem?id=2243 Name : Knight Moves Date : Tuesday, April 20, 2010 Time Stage : 0:14 Result: Test Data: //----------------------------------------------------------------------------*/ #include <iostream> #include <string.h> #include <stdio.h> #include <queue> using namespace std; struct node { int x, y; }; const int directions[8][2] = { {2, 1}, {1, 2}, {2, -1}, {-1, 2}, {1, -2}, {-2, 1}, {-1, -2}, {-2, -1} }; char a, b; int x, y, x2, y2; int minTrip, tmpTrip; int mark[10][10]; void out( ) { int i, j; for( i = 1; i <= 8; i++ ) { for( j = 1; j <= 8; j++ ) cout << mark[i][j] << " "; cout << endl; } } void bfs() { memset( mark, 0, sizeof( mark ) ); mark[x][y] = 1; if( x == x2 && y == y2 ) return ; int i, j, k; minTrip = 0x0f0f0f0f; tmpTrip = 0; node n; queue<node> q; n.x = x; n.y = y; q.push( n ); while( !q.empty() ) { n = q.front(); q.pop(); for( i = 0; i < 8; i++ ) { node nt; nt.x = n.x + directions[i][0]; nt.y = n.y + directions[i][1]; if( nt.x <= 0 || nt.y <= 0 || nt.x > 8 || nt.y > 8 ) continue; // border if( !mark[nt.x][nt.y] ) { mark[nt.x][nt.y] = mark[n.x][n.y] + 1; q.push( nt ); } if( nt.x == x2 && nt.y == y2 ) // final point { return ; } } } } int main() { int i, j, k, tcase; while( cin >> a >> y >> b >> y2 ) { x = a - 'a' + 1; x2 = b - 'a' + 1; bfs(); printf("To get from %c%d to %c%d takes %d knight moves./n", a, y, b, y2, mark[x2][y2]-1); } return 0; }