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package class_2022_01_1_week;
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import java.util.ArrayList;
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// 本题是Code02_CatAndMoused的补充code
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// leetcode官网增加了测试用例
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// 进一步说明，贪心的方式其实是没有道理的
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// 不过原来贪心的方法，把源代码第35行
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// int limit = (n << 1) + 3;
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// 改成：
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// int limit = (n << 1) + 12;
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// 依然继续能通过
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// 但是贪心依然是没道理的，人为构造的话总能找出反例来
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// 所以补充了一个拓扑排序的解法
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// 这个方法，没在课上讲，有空我们补充一下
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public class Problem_0913_CatAndMouse {
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public static final int MOUSE_TURN = 0, CAT_TURN = 1;
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public static final int DRAW = 0, MOUSE_WIN = 1, CAT_WIN = 2;
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public int catMouseGame(int[][] graph) {
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int n = graph.length;
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int[][][] indegree = new int[n][n][2];
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for (int i = 0; i < n; i++) {
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for (int j = 1; j < n; j++) {
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indegree[i][j][MOUSE_TURN] = graph[i].length;
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indegree[i][j][CAT_TURN] = graph[j].length;
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}
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}
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for (int i = 0; i < n; i++) {
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for (int node : graph[0]) {
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indegree[i][node][CAT_TURN]--;
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}
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}
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int[][][] ans = new int[n][n][2];
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int[][] queue = new int[n * n * 2][];
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int left = 0;
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int right = 0;
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for (int j = 1; j < n; j++) {
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ans[0][j][MOUSE_TURN] = MOUSE_WIN;
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ans[0][j][CAT_TURN] = MOUSE_WIN;
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queue[right++] = new int[] { 0, j, MOUSE_TURN };
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queue[right++] = new int[] { 0, j, CAT_TURN };
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}
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for (int i = 1; i < n; i++) {
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ans[i][i][MOUSE_TURN] = CAT_WIN;
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ans[i][i][CAT_TURN] = CAT_WIN;
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queue[right++] = new int[] { i, i, MOUSE_TURN };
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queue[right++] = new int[] { i, i, CAT_TURN };
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}
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while (left != right) {
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int[] cur = queue[left++];
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int mouse = cur[0], cat = cur[1], turn = cur[2];
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int curAns = ans[mouse][cat][turn];
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for (int[] prevState : preStatus(graph, mouse, cat, turn)) {
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int prevMouse = prevState[0], prevCat = prevState[1], prevTurn = prevState[2];
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if (ans[prevMouse][prevCat][prevTurn] == DRAW) {
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boolean canWin = (curAns == MOUSE_WIN && prevTurn == MOUSE_TURN)
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|| (curAns == CAT_WIN && prevTurn == CAT_TURN);
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if (canWin) {
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ans[prevMouse][prevCat][prevTurn] = curAns;
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queue[right++] = new int[] { prevMouse, prevCat, prevTurn };
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} else {
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if (--indegree[prevMouse][prevCat][prevTurn] == 0) {
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int lose = prevTurn == MOUSE_TURN ? CAT_WIN : MOUSE_WIN;
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ans[prevMouse][prevCat][prevTurn] = lose;
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queue[right++] = new int[] { prevMouse, prevCat, prevTurn };
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}
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}
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}
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}
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}
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return ans[1][2][MOUSE_TURN];
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}
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public ArrayList<int[]> preStatus(int[][] graph, int mouse, int cat, int turn) {
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ArrayList<int[]> prevStates = new ArrayList<int[]>();
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int prevTurn = turn == MOUSE_TURN ? CAT_TURN : MOUSE_TURN;
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if (prevTurn == MOUSE_TURN) {
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for (int prev : graph[mouse]) {
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prevStates.add(new int[] { prev, cat, prevTurn });
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}
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} else {
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for (int prev : graph[cat]) {
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if (prev != 0) {
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prevStates.add(new int[] { mouse, prev, prevTurn });
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}
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}
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}
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return prevStates;
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}
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}
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