[Swift]LeetCode913.猫与老鼠 | Cat and Mouse

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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. 

---

一个无向图上的游戏由两个玩家组成，鼠标和猫，他们交替轮流。

该图如下给出：graph[a]是所有节点的列表，b例如ab图的边缘。

鼠标从节点1开始并先行，Cat从节点2开始然后变为第二个，并且节点0处有一个孔。

在每个玩家的回合中，他们必须沿着图表的一个边缘行进，以满足它们的位置。例如，如果鼠标位于节点1，则它必须前往任何节点graph[1]。

此外，Cat不允许移动到孔（节点0）。

然后，游戏可以以3种方式结束：

• 如果Cat占用与鼠标相同的节点，则Cat获胜。
• 如果鼠标到达了洞，鼠标就会获胜。
• 如果一个位置被重复（即，玩家处于与前一回合相同的位置，并且轮到同一个玩家），则该游戏是平局。

给定a graph，并假设两个玩家都玩得最佳，1 如果游戏是由鼠标赢得，2 如果游戏是由Cat赢得，并且0 游戏是平局，则返回。

例1：

输入：[[2,5]，[3]，[0,4,5]，[1,4,5]，[2,3]，[0,2,3]] 
输出：0
 说明：
 4- --3 --- 1
| |
2 --- 5
 \ / \
  0

注意：

1. 3 <= graph.length <= 50
2. 保证graph[1]非空。
3. 保证graph[2]包含非零元素。 

---

140ms

class Solution {
    func catMouseGame(_ graph: [[Int]]) -> Int {
        var n:Int = graph.count
        var win:[[Int]] = [[Int]](repeating: [Int](repeating: 0, count: n*n), count: 2)
        //mc
        for i in 0..<n
        {
            win[0][i] = 1
            win[1][i] = 1
        }
        for i in 0..<n
        {
            win[0][i*n+i] = 2
            win[1][i*n+i] = 2
        }
        
        while(true)
        {
            var anew:Bool = false
            for m in 0..<n
            {
                inner:
                for c in 1..<n
                {
                    if win[0][m*n+c] == 0
                    {
                        var und:Bool = false
                        for e in graph[m]
                        {
                            if win[1][e*n+c] == 1
                            {
                                win[0][m*n+c] = 1
                                anew = true
                                continue inner
                            }
                            if win[1][e*n+c] == 0
                            {
                                    und = true
                            }
                        }
                        if !und
                        {
                            win[0][m*n+c] = 2
                            anew = true
                        }
                    }
                }
            }
            for c in 1..<n
            {
                inner:
                for m in 0..<n
                {
                    if win[1][m*n+c] == 0
                    {
                        var und:Bool = false
                         for e in graph[c]
                        {
                            if e == 0 {continue}
                            if win[0][m*n+e] == 2
                            {
                                win[1][m*n+c] = 2
                                anew = true
                                continue inner
                            }
                            if win[0][m*n+e] == 0
                            {
                                und = true
                            }
                        }
                        if !und
                        {
                             win[1][m*n+c] = 1
                            anew = true
                        }
                    }
                }
            }
            if !anew {break}
        }
        return win[0][1*n+2]
    }
}