#P2208. [USACO13OPEN] What's Up With Gravity S

    ID: 1225 远端评测题 1000ms 125MiB 尝试: 0 已通过: 0 难度: (无) 上传者: 标签>分治图论广度优先搜索BFS枚举暴力模拟USACO2013

[USACO13OPEN] What's Up With Gravity S

题目描述

Captain Bovidian is on an adventure to rescue her crew member, Doctor Beefalo. Like all great adventures, this story plays out in a two dimensional N by M grid (1 <= N, M <= 500), representing a side view of the captain's world. Some grid cells are empty while others are blocked and cannot be traversed.

Unfortunately, Captain Bovidian cannot jump. She must obey the following rules of physics while traversing her world.

  1. If there is no cell directly underneath Captain Bovidian (that is, if she is at the edge of the grid), then she flies out into space and fails her mission.

  2. If the cell directly underneath Captain Bovidian is empty, then she falls into that cell.

  3. Otherwise:

a) Captain Bovidian may move left or right if the corresponding cell exists and is empty.

b) Or, Captain Bovidian may flip the direction of gravity.

When Captain Bovidian changes the direction of gravity, the cell that's 'underneath' her (as mentioned in rules 1 and 2) toggles between the cell with one higher row index and the cell with one lower row index (the first row in the input has index 1, and the last row has index N). Initially, the cells with one higher row index are underneath Captain Bovidian.

Doctor Beefalo is lost somewhere in this world. Help Captain Bovidian arrive at her cell using the least number of gravity flips as possible. If it is impossible to reach Doctor Beefalo, please output -1.

Bovidian 船长正在拯救她的船员,Beefalo 博士。

和所有伟大的冒险故事一样,这个故事也是发生在一个2D平面上的。囧

这个平面是M*N的格子组成的网格,代表着船长的世界的一个侧视图。

有些格子是空的,另一些则是实心的,并且不能直接通过。

很不幸的是,船长跳不起来。她必须遵守这个世界的特殊物理法则。

1)如果船长的正下方没有方块(换句话说,即使她正在网格的边缘),那么她就会掉入宇宙中,同时意味着冒险失败。

2)如果船长的正下方的方块是空的,那么她就会掉到这个方块,

3)在不满足1)与2)的情况下,船长可以做一下的事情:

a) 如果左边(或右边)的方格是空的,那么她可以走到那个格子。

b船长可以翻转重力的方向

当船长改变翻转重力的方向时,我们就改变船长”下方“的定义。

”下方“的定义只能是两种

(A)比船长位置所在的方格的列编号更大的格子,

(B)比船长位置所在的方格的列编号更小的格子,

一开始的时候,“下方”的定义是比船长位置所在方格的列编号更大的格子。

Beefalo博士正迷失在这个世界中的某一处,请帮助船长从起点到达博士的地方。

如果可以到达,请输出最少需要的翻转重力的次数。

如果不可以到达,请输出-1

输入格式

第1行输入两个整数 N,M

第2行到N+1行中,第i+1行则是代表船长世界的第i行。每行有M个字符。

其中","代表着一个空的格子,"#"代表着一个实心的格子,"C"代表着船长的位置,"D"代表着博士的位置。

输出格式

一行,输出一个整数。

如果船长可以到达,请输出最少需要的翻转重力的次数。

如果不可以到达,请输出-1

5 5
#####
#...#
#...D
#C...
##.##
3

提示

输出解释:

首先,船长在(4,2),接着她翻转重力,到达(2,2)

接着她向右走走到(2,4),接着她第二次翻转重力,到达(4,4)

然后她继续向右走到(4,5),最后在翻转一次重力,到达博士所在的(3,5)