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feat: add solutions to lc problem: No.0913
No.0913.Cat and Mouse
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‎solution/0900-0999/0913.Cat and Mouse/README.md

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@@ -64,15 +64,267 @@
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def catMouseGame(self, graph: List[List[int]]) -> int:
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@lru_cache(None)
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def dfs(i, j, k):
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# 老鼠 / 猫 / 总步数
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if k >= 2 * len(graph):
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return 0 # 平局
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if i == 0:
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return 1 # 老鼠获胜
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if i == j:
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return 2 # 猫获胜
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if k % 2: # 轮到猫行动
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tie = False
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for next in graph[j]:
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if next == 0:
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continue
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x = dfs(i, next, k + 1)
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if x == 2:
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return 2
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if x == 0:
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# 继续寻找是否存在获胜的方案
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tie = True
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if tie:
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return 0
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return 1
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else: # 轮到老鼠行动
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tie = False
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for next in graph[i]:
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x = dfs(next, j, k + 1)
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if x == 1:
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return 1
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if x == 0:
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# 继续寻找是否存在获胜的方案
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tie = True
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if tie:
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return 0
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return 2
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return dfs(1, 2, 0)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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private int[][][] memo;
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private int[][] graph;
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public int catMouseGame(int[][] graph) {
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int n = graph.length;
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this.graph = graph;
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memo = new int[n][n][2 * n + 10];
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for (int i = 0; i < n; ++i) {
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for (int j = 0; j < n; ++j) {
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for (int k = 0; k < 2 * n + 10; ++k) {
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memo[i][j][k] = -1;
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}
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}
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}
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return dfs(1, 2, 0);
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}
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private int dfs(int i, int j, int k) {
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// 老鼠 / 猫 / 总步数
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if (memo[i][j][k] != -1) {
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return memo[i][j][k];
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}
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if (k >= 2 * graph.length) {
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// 平局
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memo[i][j][k] = 0;
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} else if (i == 0) {
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// 老鼠获胜
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memo[i][j][k] = 1;
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} else if (i == j) {
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// 猫获胜
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memo[i][j][k] = 2;
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} else if (k % 2 == 1) {
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// 轮到猫行动
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boolean tie = false;
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boolean win = false;
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for (int next : graph[j]) {
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if (next == 0) {
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continue;
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}
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int x = dfs(i, next, k + 1);
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if (x == 2) {
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win = true;
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memo[i][j][k] = 2;
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break;
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}
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if (x == 0) {
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// 继续寻找是否存在获胜的方案
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tie = true;
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}
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}
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if (!win) {
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memo[i][j][k] = tie ? 0 : 1;
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}
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} else {
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// 轮到老鼠行动
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boolean tie = false;
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boolean win = false;
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for (int next : graph[i]) {
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int x = dfs(next, j, k + 1);
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if (x == 1) {
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win = true;
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memo[i][j][k] = 1;
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break;
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}
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if (x == 0) {
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// 继续寻找是否存在获胜的方案
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tie = true;
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}
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}
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if (!win) {
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memo[i][j][k] = tie ? 0 : 2;
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}
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}
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return memo[i][j][k];
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<vector<vector<int>>> memo;
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vector<vector<int>> graph;
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int catMouseGame(vector<vector<int>>& graph) {
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int n = graph.size();
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this->graph = graph;
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memo.resize(n, vector<vector<int>>(n, vector<int>(2 * n + 10)));
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for (int i = 0; i < memo.size(); ++i)
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for (int j = 0; j < memo[0].size(); ++j)
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for (int k = 0; k < memo[0][0].size(); ++k)
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memo[i][j][k] = -1;
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return dfs(1, 2, 0);
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}
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int dfs(int i, int j, int k) {
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if (memo[i][j][k] != -1) return memo[i][j][k];
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if (k >= 2 * graph.size()) memo[i][j][k] = 0;
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else if (i == 0) memo[i][j][k] = 1;
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else if (i == j) memo[i][j][k] = 2;
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else if (k % 2)
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{
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bool tie = false, win = false;
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for (int next : graph[j])
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{
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if (next == 0) continue;
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int x = dfs(i, next, k + 1);
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if (x == 2)
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{
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win = true;
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memo[i][j][k] = 2;
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break;
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}
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if (x == 0) tie = true;
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}
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if (!win) memo[i][j][k] = tie ? 0 : 1;
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}
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else
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{
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bool tie = false, win = false;
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for (int next : graph[i])
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{
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int x = dfs(next, j, k + 1);
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if (x == 1)
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{
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win = true;
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memo[i][j][k] = 1;
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break;
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}
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if (x == 0) tie = true;
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}
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if (!win) memo[i][j][k] = tie ? 0 : 2;
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}
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return memo[i][j][k];
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}
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};
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```
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### **Go**
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```go
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func catMouseGame(graph [][]int) int {
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n := len(graph)
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memo := make([][][]int, n)
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for i := range memo {
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memo[i] = make([][]int, n)
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for j := range memo[i] {
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memo[i][j] = make([]int, 2*n+10)
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for k := range memo[i][j] {
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memo[i][j][k] = -1
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}
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}
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}
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var dfs func(i, j, k int) int
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dfs = func(i, j, k int) int {
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if memo[i][j][k] != -1 {
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return memo[i][j][k]
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}
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if k >= 2*len(graph) {
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memo[i][j][k] = 0
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} else if i == 0 {
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memo[i][j][k] = 1
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} else if i == j {
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memo[i][j][k] = 2
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} else if k%2 == 1 {
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tie, win := false, false
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for _, next := range graph[j] {
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if next == 0 {
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continue
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}
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x := dfs(i, next, k+1)
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if x == 2 {
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win = true
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memo[i][j][k] = 2
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break
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}
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if x == 0 {
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tie = true
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}
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}
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if !win {
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if tie {
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memo[i][j][k] = 0
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} else {
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memo[i][j][k] = 1
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}
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}
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} else {
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tie, win := false, false
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for _, next := range graph[i] {
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x := dfs(next, j, k+1)
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if x == 1 {
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win = true
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memo[i][j][k] = 1
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break
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}
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if x == 0 {
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tie = true
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}
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}
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if !win {
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if tie {
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memo[i][j][k] = 0
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} else {
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memo[i][j][k] = 2
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}
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}
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}
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return memo[i][j][k]
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}
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return dfs(1, 2, 0)
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}
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```
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### **...**

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