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package class_2022_05_2_week;
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import java.util.Arrays;
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// 来自字节
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// 5.6笔试
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// 给定N件物品，每个物品有重量(w[i])、有价值(v[i])
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// 只能最多选两件商品，重量不超过bag，返回价值最大能是多少？
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// N <= 10^5, w[i] <= 10^5, v[i] <= 10^5, bag <= 10^5
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// 本题的关键点：什么数据范围都很大，唯独只需要最多选两件商品，这个可以利用一下
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public class Code01_TwoObjectMaxValue {
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// 暴力方法
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// 为了验证而写的方法
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public static int max1(int[] w, int[] v, int bag) {
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return process1(w, v, 0, 2, bag);
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}
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public static int process1(int[] w, int[] v, int index, int restNumber, int restWeight) {
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if (restNumber < 0 || restWeight < 0) {
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return -1;
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}
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if (index == w.length) {
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return 0;
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}
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int p1 = process1(w, v, index + 1, restNumber, restWeight);
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int p2 = -1;
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int next = process1(w, v, index + 1, restNumber - 1, restWeight - w[index]);
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if (next != -1) {
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p2 = v[index] + next;
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}
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return Math.max(p1, p2);
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}
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// 正式方法
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// 时间复杂度O(N * logN)
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public static int max2(int[] w, int[] v, int bag) {
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int n = w.length;
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int[][] arr = new int[n][2];
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for (int i = 0; i < n; i++) {
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arr[i][0] = w[i];
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arr[i][1] = v[i];
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}
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// O(N * logN)
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Arrays.sort(arr, (a, b) -> (a[0] - b[0]));
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// 重量从轻到重，依次标号1、2、3、4....
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// 价值依次被构建成了RMQ结构
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// O(N * logN)
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RMQ rmq = new RMQ(arr);
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int ans = 0;
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// N * logN
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for (int i = 0, j = 1; i < n && arr[i][0] <= bag; i++, j++) {
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// 当前来到0号货物，RMQ结构1号
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// 当前来到i号货物，RMQ结构i+1号
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// 查询重量的边界，重量 边界 <= bag - 当前货物的重量
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// 货物数组中，找到 <= 边界，最右的位置i
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// RMQ，位置 i + 1
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int right = right(arr, bag - arr[i][0]) + 1;
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int rest = 0;
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// j == i + 1，当前的货物，在RMQ里的下标
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if (right == j) {
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rest = rmq.max(1, right - 1);
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} else if (right < j) {
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rest = rmq.max(1, right);
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} else { // right > j
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rest = Math.max(rmq.max(1, j - 1), rmq.max(j + 1, right));
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}
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ans = Math.max(ans, arr[i][1] + rest);
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}
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return ans;
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}
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public static int right(int[][] arr, int limit) {
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int l = 0;
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int r = arr.length - 1;
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int m = 0;
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int ans = -1;
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while (l <= r) {
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m = (l + r) / 2;
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if (arr[m][0] <= limit) {
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ans = m;
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l = m + 1;
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} else {
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r = m - 1;
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}
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}
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return ans;
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}
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public static class RMQ {
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public int[][] max;
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public RMQ(int[][] arr) {
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int n = arr.length;
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int k = power2(n);
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max = new int[n + 1][k + 1];
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for (int i = 1; i <= n; i++) {
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max[i][0] = arr[i - 1][1];
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}
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for (int j = 1; (1 << j) <= n; j++) {
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for (int i = 1; i + (1 << j) - 1 <= n; i++) {
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max[i][j] = Math.max(max[i][j - 1], max[i + (1 << (j - 1))][j - 1]);
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}
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}
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}
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public int max(int l, int r) {
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if (r < l) {
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return 0;
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}
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int k = power2(r - l + 1);
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return Math.max(max[l][k], max[r - (1 << k) + 1][k]);
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}
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private int power2(int m) {
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int ans = 0;
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while ((1 << ans) <= (m >> 1)) {
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ans++;
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}
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return ans;
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}
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}
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// 为了测试
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public static int[] randomArray(int n, int v) {
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int[] arr = new int[n];
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for (int i = 0; i < n; i++) {
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arr[i] = (int) (Math.random() * v);
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}
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return arr;
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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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int N = 12;
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int V = 20;
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int testTimes = 5000;
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System.out.println("测试开始");
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for (int i = 0; i < testTimes; i++) {
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int n = (int) (Math.random() * N) + 1;
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int[] w = randomArray(n, V);
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int[] v = randomArray(n, V);
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int bag = (int) (Math.random() * V * 3);
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int ans1 = max1(w, v, bag);
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int ans2 = max2(w, v, bag);
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if (ans1 != ans2) {
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System.out.println("出错了!");
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break;
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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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