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;
}