POJ 2243 Knight Moves

bfs,使用C++的queue，300ms。

# include <cstdio>

# include <cstring>

# include <queue>



using namespace std;



const int dx[] = {1,1,-1,-1,2,2,-2,-2};

const int dy[] = {2,-2,2,-2,1,-1,1,-1};



struct Pos{int x, y, d;};



char s[5], g[5];

int sx, sy, gx, gy;



int bfs(void)

{

    queue <Pos> Q;

    Pos cur, nst;

    char vis[9][9];



    memset(vis, 0, sizeof(vis));



    vis[sx][sy] = 1;

    cur.x = sx, cur.y = sy, cur.d = 0;

    Q.push(cur);

    while (!Q.empty())

    {

        cur = Q.front(); Q.pop();

        if (cur.x==gx && cur.y==gy) return cur.d;

        for (int i = 0; i < 8; ++i)

        {

            nst.x = cur.x + dx[i];

            nst.y = cur.y + dy[i];

            if (1<=nst.x&&nst.x<=8 && 1<=nst.y&&nst.y<=8 && !vis[nst.x][nst.y])

            {

                if (nst.x==gx && nst.y==gy) return cur.d+1;

                nst.d = cur.d + 1;

                vis[nst.x][nst.y] = 1;

                Q.push(nst);

            }

        }

    }



//    return -1;

}



void solve(void)

{

    int ans = bfs();

    printf("To get from %s to %s takes %d knight moves.\n", s, g, ans);

}



int main()

{

    while (~scanf("%s%s", s, g))

    {

        sx = s[0]-'a'+1;

        sy = s[1]-'1'+1;

        gx = g[0]-'a'+1;

        gy = g[1]-'1'+1;

        solve();

    }



    return 0;

}