A game on an undirected graph is played by two players, Mouse and Cat, who alternate turns.
The graph is given as follows: graph[a] is a list of all nodes b such that ab is an edge of the graph.
Mouse starts at node 1 and goes first, Cat starts at node 2 and goes second, and there is a Hole at node 0.
During each player's turn, they must travel along one edge of the graph that meets where they are. For example, if the Mouse is at node 1, it must travel to any node in graph[1].
Additionally, it is not allowed for the Cat to travel to the Hole (node 0.)
Then, the game can end in 3 ways:
- If ever the Cat occupies the same node as the Mouse, the Cat wins.
- If ever the Mouse reaches the Hole, the Mouse wins.
- If ever a position is repeated (ie. the players are in the same position as a previous turn, and it is the same player's turn to move), the game is a draw.
Given a graph, and assuming both players play optimally, return 1 if the game is won by Mouse, 2 if the game is won by Cat, and 0 if the game is a draw.
Example 1:
Input: [[2,5],[3],[0,4,5],[1,4,5],[2,3],[0,2,3]] Output: 0 Explanation: 4---3---1 | | 2---5 / 0
Note:
3 <= graph.length <= 50- It is guaranteed that
graph[1]is non-empty. - It is guaranteed that
graph[2]contains a non-zero element.
思路:模拟一遍游戏,
1. 注意返回值,如果你设定的返回是一个数,就照着一个数的思想去想
2. 在求某个状态的时候,先把这个状态设置为draw,这样可以处理环的问题
https://leetcode.com/problems/cat-and-mouse/discuss/175892/recursion-+-memorization-84ms-Python
class Solution(object):
def catMouseGame(self, graph):
"""
:type graph: List[List[int]]
:rtype: int
"""
mouse,cat={},{}
def mouse_move(i,j):
if (i,j) in mouse: return mouse[(i,j)]
mouse[(i,j)]=0
canDraw = False
for t in graph[i]:
if t==0:
mouse[(i,j)]=1
return 1
if t==j: continue
tmp = cat_move(t,j)
if tmp==1:
mouse[(i,j)]=1
return 1
if tmp==0: canDraw=True
res = 0 if canDraw else 2
mouse[(i,j)]=res
return res
def cat_move(i,j):
if (i,j) in cat: return cat[(i,j)]
cat[(i,j)]=0 # handle draw case
canDraw = False
for t in graph[i]:
# case1
if t==j:
cat[(i,j)]=2
return 2
# case2
if t==0: continue
# case3
tmp = mouse_move(i,t)
if tmp==2:
cat[(i,j)]=2
return 2
if tmp==0: canDraw=True
res = 0 if canDraw else 1
cat[(i,j)]=res
return res
return mouse_move(1,2)
s=Solution()
print(s.catMouseGame([[2,5],[3],[0,4,5],[1,4,5],[2,3],[0,2,3]]))
最后
以上就是单纯绿草最近收集整理的关于913. Cat and Mouse的全部内容,更多相关913.内容请搜索靠谱客的其他文章。
发表评论 取消回复