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package class_2022_05_1_week;
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// 来自微众
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// 人工智能岗
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// 一开始有21个球，甲和乙轮流拿球，甲先、乙后
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// 每个人在自己的回合，一定要拿不超过3个球，不能不拿
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// 最终谁的总球数为偶数，谁赢
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// 请问谁有必胜策略
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public class Code02_WhoWin21Balls {
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// balls = 21
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// ball是奇数
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public static String win(int balls) {
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return process(0, balls, 0, 0);
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}
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// 憋递归！
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// turn 谁的回合！
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// turn == 0 甲回合
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// turn == 1 乙回合
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// rest剩余球的数量
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// 之前，jiaBalls、yiBalls告诉你！
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// 当前，根据turn，知道是谁的回合！
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// 当前，还剩多少球，rest
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// 返回：谁会赢！
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public static String process(int turn, int rest, int jia, int yi) {
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if (rest == 0) {
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return (jia & 1) == 0 ? "甲" : "乙";
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}
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// rest > 0, 还剩下球！
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if (turn == 0) { // 甲的回合！
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// 甲，自己赢！甲赢！
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for (int pick = 1; pick <= Math.min(rest, 3); pick++) {
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// pick 甲当前做的选择
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if (process(1, rest - pick, jia + pick, yi).equals("甲")) {
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return "甲";
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}
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}
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return "乙";
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} else {
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for (int pick = 1; pick <= Math.min(rest, 3); pick++) {
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// pick 甲当前做的选择
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if (process(0, rest - pick, jia, yi + pick).equals("乙")) {
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return "乙";
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}
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}
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return "甲";
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}
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}
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// 我们补充一下设定，假设一开始的球数量不是21，是任意的正数
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// 如果最终两个人拿的都是偶数，认为无人获胜，平局
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// 如果最终两个人拿的都是奇数，认为无人获胜，平局
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// rest代表目前剩下多少球
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// cur == 0 代表目前是甲行动
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// cur == 1 代表目前是乙行动
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// first == 0 代表目前甲所选的球数，加起来是偶数
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// first == 1 代表目前甲所选的球数，加起来是奇数
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// second == 0 代表目前乙所选的球数，加起来是偶数
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// second == 1 代表目前乙所选的球数，加起来是奇数
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// 返回选完了rest个球，谁会赢，只会返回"甲"、"乙"、"平"
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// win1方法，就是彻底暴力的做所有尝试，并且返回最终的胜利者
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// 在甲的回合，甲会尝试所有的可能，以保证自己会赢，如果自己怎么都不会赢，那也要尽量平局，如果这个也不行，只能对方赢
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// 在乙的回合，乙会尝试所有的可能，以保证自己会赢，如果自己怎么都不会赢，那也要尽量平局，如果这个也不行，只能对方赢
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// 算法和数据结构体系学习班，视频39章节，牛羊吃草问题，就是类似这种递归
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public static String win1(int rest, int cur, int first, int second) {
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if (rest == 0) {
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if (first == 0 && second == 1) {
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return "甲";
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}
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if (first == 1 && second == 0) {
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return "乙";
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}
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return "平";
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}
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if (cur == 0) { // 甲行动
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String bestAns = "乙";
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for (int pick = 1; pick <= Math.min(3, rest); pick++) {
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String curAns = win1(rest - pick, 1, first ^ (pick & 1), second);
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if (curAns.equals("甲")) {
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bestAns = "甲";
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break;
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}
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if (curAns.equals("平")) {
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bestAns = "平";
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}
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}
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return bestAns;
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} else { // 乙行动
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String bestAns = "甲";
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for (int pick = 1; pick <= Math.min(3, rest); pick++) {
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String curAns = win1(rest - pick, 0, first, second ^ (pick & 1));
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if (curAns.equals("乙")) {
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bestAns = "乙";
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break;
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}
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if (curAns.equals("平")) {
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bestAns = "平";
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}
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}
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return bestAns;
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}
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}
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// 下面的win2方法，仅仅是把win1方法，做了记忆化搜索
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// 变成了动态规划
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public static String[][][][] dp = new String[5000][2][2][2];
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public static String win2(int rest, int cur, int first, int second) {
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if (rest == 0) {
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if (first == 0 && second == 1) {
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return "甲";
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}
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if (first == 1 && second == 0) {
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return "乙";
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}
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return "平";
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}
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if (dp[rest][cur][first][second] != null) {
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return dp[rest][cur][first][second];
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}
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if (cur == 0) { // 甲行动
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String bestAns = "乙";
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for (int pick = 1; pick <= Math.min(3, rest); pick++) {
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String curAns = win2(rest - pick, 1, first ^ (pick & 1), second);
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if (curAns.equals("甲")) {
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bestAns = "甲";
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break;
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}
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if (curAns.equals("平")) {
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bestAns = "平";
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}
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}
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dp[rest][cur][first][second] = bestAns;
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return bestAns;
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} else { // 乙行动
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String bestAns = "甲";
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for (int pick = 1; pick <= Math.min(3, rest); pick++) {
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String curAns = win2(rest - pick, 0, first, second ^ (pick & 1));
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if (curAns.equals("乙")) {
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bestAns = "乙";
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break;
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}
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if (curAns.equals("平")) {
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bestAns = "平";
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}
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}
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dp[rest][cur][first][second] = bestAns;
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return bestAns;
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}
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}
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// 为了测试
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public static void main(String[] args) {
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for (int balls = 1; balls <= 500; balls += 2) {
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System.out.println("球数为 " + balls + " 时 , 赢的是 " + win(balls));
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}
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}
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} |
