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package class02;
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import java.util.Arrays;
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public class Code04_Drive {
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// 课上的现场版本
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// income -> N * 2 的矩阵 N是偶数！
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// 0 [9, 13]
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// 1 [45,60]
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public static int maxMoney1(int[][] income) {
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if (income == null || income.length < 2 || (income.length & 1) != 0) {
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return 0;
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}
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int N = income.length; // 司机数量一定是偶数，所以才能平分，A N /2 B N/2
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int M = N >> 1; // M = N / 2 要去A区域的人
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return process1(income, 0, M);
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}
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// index.....所有的司机，往A和B区域分配！
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// A区域还有rest个名额!
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// 返回把index...司机，分配完，并且最终A和B区域同样多的情况下，index...这些司机，整体收入最大是多少！
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public static int process1(int[][] income, int index, int rest) {
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if (index == income.length) {
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return 0;
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}
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// 还剩下司机！
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if (income.length - index == rest) {
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return income[index][0] + process1(income, index + 1, rest - 1);
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}
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if (rest == 0) {
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return income[index][1] + process1(income, index + 1, rest);
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}
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// 当前司机，可以去A，或者去B
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int p1 = income[index][0] + process1(income, index + 1, rest - 1);
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int p2 = income[index][1] + process1(income, index + 1, rest);
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return Math.max(p1, p2);
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}
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// 严格位置依赖的动态规划版本
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public static int maxMoney2(int[][] income) {
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int N = income.length;
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int M = N >> 1;
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int[][] dp = new int[N + 1][M + 1];
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for (int i = N - 1; i >= 0; i--) {
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for (int j = 0; j <= M; j++) {
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if (N - i == j) {
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dp[i][j] = income[i][0] + dp[i + 1][j - 1];
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} else if (j == 0) {
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dp[i][j] = income[i][1] + dp[i + 1][j];
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} else {
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int p1 = income[i][0] + dp[i + 1][j - 1];
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int p2 = income[i][1] + dp[i + 1][j];
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dp[i][j] = Math.max(p1, p2);
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}
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}
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}
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return dp[0][M];
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}
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// 这题有贪心策略 :
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// 假设一共有10个司机，思路是先让所有司机去A，得到一个总收益
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// 然后看看哪5个司机改换门庭(去B)，可以获得最大的额外收益
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// 这道题有贪心策略，打了我的脸
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// 但是我课上提到的技巧请大家重视
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// 根据数据量猜解法可以省去大量的多余分析，节省时间
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// 这里感谢卢圣文同学
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public static int maxMoney3(int[][] income) {
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int N = income.length;
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int[] arr = new int[N];
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int sum = 0;
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for (int i = 0; i < N; i++) {
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arr[i] = income[i][1] - income[i][0];
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sum += income[i][0];
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}
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Arrays.sort(arr);
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int M = N >> 1;
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for (int i = N - 1; i >= M; i--) {
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sum += arr[i];
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}
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return sum;
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}
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// 找到了leetcode上的测试
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// leetcode上让求最小，课上讲的求最大
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// 其实是一个意思
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// 测试链接 : https://leetcode.cn/problems/two-city-scheduling/
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public static int twoCitySchedCost(int[][] costs) {
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int N = costs.length;
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int[] arr = new int[N];
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int sum = 0;
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for (int i = 0; i < N; i++) {
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arr[i] = costs[i][1] - costs[i][0];
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sum += costs[i][0];
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}
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Arrays.sort(arr);
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int M = N >> 1;
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for (int i = 0; i < M; i++) {
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sum += arr[i];
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}
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return sum;
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}
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// 返回随机len*2大小的正数矩阵
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// 值在0~value-1之间
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public static int[][] randomMatrix(int len, int value) {
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int[][] ans = new int[len << 1][2];
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for (int i = 0; i < ans.length; i++) {
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ans[i][0] = (int) (Math.random() * value);
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ans[i][1] = (int) (Math.random() * value);
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}
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return ans;
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}
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public static void main(String[] args) {
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int N = 10;
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int value = 100;
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int testTime = 500;
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System.out.println("测试开始");
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for (int i = 0; i < testTime; i++) {
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int len = (int) (Math.random() * N) + 1;
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int[][] matrix = randomMatrix(len, value);
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int ans1 = maxMoney1(matrix);
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int ans2 = maxMoney2(matrix);
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int ans3 = maxMoney3(matrix);
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if (ans1 != ans2 || ans1 != ans3) {
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System.out.println(ans1);
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System.out.println(ans2);
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System.out.println(ans3);
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System.out.println("Oops!");
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}
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}
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System.out.println("测试结束");
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}
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}
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