【LeetCode】913. Cat and Mouse 解题报告（Python）

作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

---

目录

• • 题目描述
  • 题目大意
  • 解题方法
  • 参考资料
  • 日期

题目地址：https://leetcode.com/problems/cat-and-mouse/description/

题目描述

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:

1. 3 <= graph.length <= 50
2. It is guaranteed that graph[1] is non-empty.
3. It is guaranteed that graph[2] contains a non-zero element.

题目大意

猫鼠游戏。

有一张无向图，包含最多50个结点。有两个玩家（Mouse和Cat）在图上，Mouse的起始位置是1，Cat的起始位置是2，0处有一个洞，Cat不能移动到0。Mouse和Cat在图上轮流移动，每次必须移动到与当前结点相邻的一个结点。

游戏有三种结束的可能：

1. Cat和Mouse进入同一结点，则Cat胜利
2. Mouse进入0结点，则Mouse胜利
3. Cat和Mouse的位置和回合发生了重复，则平局

问：如果Cat和Mouse都以最优策略行动，最后的结果是什么？

解题方法

这个题实在太难了，我只能参考别人的做法了。正确的做法应该是BFS，而且是已知结果倒着求过程的BFS。

设计节点状态是(m,c,turn)，用color[m][c][turn]来记忆该状态的输赢情况．

首先我们将所有已知的确定状态加入一个队列．已知状态包括(0,c,turn)肯定是老鼠赢，(x,x,turn)且x!=0肯定是猫赢。我们尝试用BFS的思路，将这些已知状态向外扩展开去．

扩展的思路是：从队列中取出队首节点状态（m,c,t），找到它的所有邻接的parent的状态（m2,c2,t2）．这里的父子关系是指，(m2,c2,t2)通过t2轮（老鼠或猫）的操作，能得到(m,c,t).我们发现，如果(m,c,t)是老鼠赢而且t2是老鼠轮，那么这个(m2,c2,t2)一定也是老鼠赢．同理，猫赢的状态也类似．于是，我们找到了一种向外扩展的方法．

向外扩展的第二个思路是：对于(m2,c2,t2)，我们再去查询它的所有children（必定是对手轮）是否都已经标注了赢的状态．如果都是赢的状态，那么说明(m2,c2,t2)无路可走，只能标记为输的状态．特别注意的是，第一条规则通过child找parent，和第二条规则通过parent找child的算法细节是不一样的，一定要小心．

这样我们通过BFS不断加入新的探明输赢的节点．直到队列为空，依然没有探明输赢的节点状态，就是平局的意思！

最后输出(1, 2, MOUSE)的颜色。没有被染过色说明是平局。

时间复杂度是O(VE)，空间复杂度是O(V).

class Solution(object):
    def catMouseGame(self, graph):
        """
        :type graph: List[List[int]]
        :rtype: int
        """
        N = len(graph)
        MOUSE, CAT = 1, 2
        # mouse, cat, turn
        color = [[[0] * 3 for i in range(N)] for j in range(N)]
        q = collections.deque()
        for i in range(1, N):
            for t in range(1, 3):
                color[0][i][t] = 1
                q.append((0, i, t))
                color[i][i][t] = 2
                q.append((i, i, t))
        while q:
            curStatus = q.popleft()
            cat, mouse, turn = curStatus
            for preStatus in self.findAllPrevStatus(graph, curStatus):
                preCat, preMouse, preTurn = preStatus
                if color[preCat][preMouse][preTurn] != 0:
                    continue
                if color[cat][mouse][turn] == 3 - turn:
                    color[preCat][preMouse][preTurn] = preTurn
                    q.append(preStatus)
                elif self.allNeighboursWin(color, graph, preStatus):
                    color[preCat][preMouse][preTurn] = 3 - preTurn
                    q.append(preStatus)
        return color[1][2][1]

    def findAllPrevStatus(self, graph, curStatus):
        ret = []
        mouse, cat, turn = curStatus
        if turn == 1:
            for preCat in graph[cat]:
                if preCat == 0:
                    continue
                ret.append((mouse, preCat, 2))
        else:
            for preMouse in graph[mouse]:
                ret.append((preMouse, cat, 1))
        return ret

    def allNeighboursWin(self, color, graph, status):
        mouse, cat, turn = status
        if turn == 1:
            for nextMouse in graph[mouse]:
                if color[nextMouse][cat][2] != 2:
                    return False
        elif turn == 2:
            for nextCat in graph[cat]:
                if nextCat == 0:
                    continue
                if color[mouse][nextCat][1] != 1:
                    return False
        return True

参考资料

https://zhanghuimeng.github.io/post/leetcode-913-cat-and-mouse/
https://github.com/wisdompeak/LeetCode/tree/master/BFS/913.Cat-and-Mouse

日期

2018 年 10 月 24 日 —— 程序员节被严重炒作了啊