Knight Moves

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BFS

题意：给你棋盘上两点的坐标，按照中国象棋中马的走法，求出其最短的步数。

// File Name: 1073.cpp// Author: bo_jwolf// Created Time: 2014年02月03日 星期一 21时24分46秒#include<vector>#include<list>#include<map>#include<set>#include<deque>#include<stack>#include<bitset>#include<algorithm>#include<functional>#include<numeric>#include<utility>#include<sstream>#include<iostream>#include<iomanip>#include<cstdio>#include<cmath>#include<cstdlib>#include<cstring>#include<ctime>#include<queue>using namespace std;int dir[ 15 ][ 5 ] = { { -2, 1 }, { -1, 2 }, { -2, -1 }, { -1, -2 },{ 2, 1 },{ 1, 2 }, { 2, -1 }, { 1, -2 } };const int maxn = 1005;int vis[ maxn ][ maxn ], sx, sy, ex, ey, Step = 0;struct node{int x, y, step;}edge[ maxn ];void Bfs(){queue< node > Q;node temp, p, q;temp.x = sx;temp.y = sy;vis[ sx ][ sy ] = 1;temp.step = 0;Q.push( temp );if( sx == ex && sy == ey ){Step = 0;return;}while( !Q.empty() ){p = Q.front();Q.pop();for( int i = 0; i < 8; ++i ){int y = p.y + dir[ i ][ 0 ];int x = p.x + dir[ i ][ 1 ];if( x >= 0 && y >= 0 && x < 8 && y < 8 && !vis[ x ][ y ] ){q.x = x;q.y = y;q.step = p.step + 1;if( x == ex && y == ey ){Step = q.step;return;}vis[ x ][ y ] = 1;Q.push( q );}}}}int main(){char str1[ maxn ], str2[ maxn ];while( scanf( "%s %s", str1, str2 ) == 2 ){sx = str1[ 0 ] - 'a';sy = str1[ 1 ] - '1';ex = str2[ 0 ] - 'a';ey = str2[ 1 ] - '1';memset( vis, 0, sizeof( vis ) );Bfs();printf( "To get from %s to %s takes %d knight moves.\n", str1, str2, Step );}return 0;}

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