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package class_2022_08_5_week;
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// 来自Amazon
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// 定义一个概念叫"变序最大和"
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// "变序最大和"是说一个数组中，每个值都可以减小或者不变，
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// 在必须把整体变成严格升序的情况下，得到的最大累加和
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// 比如，[1,100,7]变成[1,6,7]时，就有变序最大和为14
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// 比如，[5,4,9]变成[3,4,9]时，就有变序最大和为16
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// 比如，[1,4,2]变成[0,1,2]时，就有变序最大和为3
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// 给定一个数组arr，其中所有的数字都是>=0的
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// 求arr所有子数组的变序最大和中，最大的那个并返回
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// 1 <= arr长度 <= 10^6
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// 0 <= arr[i] <= 10^6
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public class Code03_SubarrayMakeSrotedMaxSum {
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// 时间复杂度O(N * V)的方法
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// 为了验证
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public static long maxSum1(int[] arr) {
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int n = arr.length;
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int max = 0;
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for (int num : arr) {
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max = Math.max(max, num);
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}
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long ans = 0;
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long[][] dp = new long[n][max + 1];
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for (int i = 0; i < n; i++) {
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for (int j = 0; j <= max; j++) {
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dp[i][j] = -1;
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}
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}
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for (int i = 0; i < n; i++) {
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ans = Math.max(ans, process1(arr, i, arr[i], dp));
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}
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return ans;
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}
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public static long process1(int[] arr, int i, int p, long[][] dp) {
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if (p <= 0 || i == -1) {
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return 0;
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}
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if (dp[i][p] != -1) {
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return dp[i][p];
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}
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int cur = Math.min(arr[i], p);
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long next = process1(arr, i - 1, cur - 1, dp);
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long ans = (long) cur + next;
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dp[i][p] = ans;
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return cur + next;
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}
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// 正式方法
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// 时间复杂度O(N)
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public static long maxSum2(int[] arr) {
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int n = arr.length;
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// 只放下标，只要有下标，arr可以拿到值
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int[] stack = new int[n];
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int size = 0;
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long[] dp = new long[n];
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long ans = 0;
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for (int i = 0; i < n; i++) {
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// i -> arr[i] 依次把收益！得到！
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int curVal = arr[i];
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int curIdx = i;
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// 20
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// 17
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while (curVal > 0 && size > 0) {
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// 100
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// 16
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int leftIdx = stack[size - 1];
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int leftVal = arr[leftIdx];
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if (leftVal >= curVal) {
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size--;
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} else {
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// leftVal < curVal
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// 8 20
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// 15 17
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int idxDiff = curIdx - leftIdx;
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int valDiff = curVal - leftVal;
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// 12 2
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// 8 19 20
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// 15 16 17
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if (valDiff >= idxDiff) {
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dp[i] += sum(curVal, idxDiff) + dp[leftIdx];
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curVal = 0;
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curIdx = 0;
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break;
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} else {
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// 18 20
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// 13 14 15 16 17
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// 17 18 19 20
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// 16
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dp[i] += sum(curVal, idxDiff);
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// 16
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// 13
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curVal -= idxDiff;
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curIdx = leftIdx;
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size--;
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}
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}
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}
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if (curVal > 0) {
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dp[i] += sum(curVal, curIdx + 1);
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}
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stack[size++] = i;
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ans = Math.max(ans, dp[i]);
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}
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return ans;
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}
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public static long sum(int max, int n) {
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n = Math.min(max, n);
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return (((long) max * 2 - n + 1) * n) / 2;
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}
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// 为了验证
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public static int[] randomArray(int n, int v) {
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int[] ans = new int[n];
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for (int i = 0; i < n; i++) {
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ans[i] = (int) (Math.random() * v);
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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 = 100;
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int V = 100;
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int testTimes = 50000;
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System.out.println("功能测试开始");
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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[] arr = randomArray(n, V);
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long ans1 = maxSum1(arr);
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long ans2 = maxSum2(arr);
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if (ans1 != ans2) {
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for (int num : arr) {
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System.out.print(num + " ");
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}
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System.out.println();
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System.out.println(ans1);
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System.out.println(ans2);
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break;
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}
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}
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System.out.println("功能测试结束");
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System.out.println("性能测试开始");
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int n = 1000000;
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int v = 1000000;
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System.out.println("数组长度 : " + n);
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System.out.println("数值范围 : " + v);
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int[] arr = randomArray(n, v);
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long start = System.currentTimeMillis();
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maxSum2(arr);
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long end = System.currentTimeMillis();
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System.out.println("运行时间 : " + (end - start) + "毫秒");
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System.out.println("性能测试结束");
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}
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}
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