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package class_2022_09_2_week;
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// 来自微众银行
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// 给定一个数字n，代表数组的长度
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// 给定一个数字m，代表数组每个位置都可以在1~m之间选择数字
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// 所有长度为n的数组中，最长递增子序列长度为3的数组，叫做达标数组
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// 返回达标数组的数量
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// 1 <= n <= 500
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// 1 <= m <= 10
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// 500 * 10 * 10 * 10
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// 结果对998244353取模
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// 实现的时候没有取模的逻辑，因为非重点
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public class Code02_NLengthMValueLIS3 {
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// 暴力方法
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// 为了验证
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public static int number1(int n, int m) {
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return process1(0, n, m, new int[n]);
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}
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public static int process1(int i, int n, int m, int[] path) {
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if (i == n) {
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return lengthOfLIS(path) == 3 ? 1 : 0;
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} else {
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int ans = 0;
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for (int cur = 1; cur <= m; cur++) {
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path[i] = cur;
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ans += process1(i + 1, n, m, path);
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}
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return ans;
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}
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}
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public static int lengthOfLIS(int[] arr) {
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if (arr == null || arr.length == 0) {
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return 0;
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}
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int[] ends = new int[arr.length];
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ends[0] = arr[0];
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int right = 0;
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int max = 1;
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for (int i = 1; i < arr.length; i++) {
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int l = 0;
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int r = right;
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while (l <= r) {
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int m = (l + r) / 2;
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if (arr[i] > ends[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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right = Math.max(right, l);
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ends[l] = arr[i];
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max = Math.max(max, l + 1);
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}
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return max;
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}
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// i : 当前来到的下标
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// f、s、t : ends数组中放置的数字！
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// ? == 0，没放！
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// n : 一共的长度！
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// m : 每一位，都可以在1~m中随意选择数字
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// 返回值：i..... 有几个合法的数组！
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public static int zuo(int i, int f, int s, int t, int n, int m) {
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if (i == n) {
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return f != 0 && s != 0 && t != 0 ? 1 : 0;
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}
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// i < n
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int ans = 0;
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for (int cur = 1; cur <= m; cur++) {
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if (f == 0 || f >= cur) {
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ans += zuo(i + 1, cur, s, t, n, m);
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} else if (s == 0 || s >= cur) {
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ans += zuo(i + 1, f, cur, t, n, m);
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} else if (t == 0 || t >= cur) {
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ans += zuo(i + 1, f, s, cur, n, m);
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}
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}
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return ans;
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}
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// 正式方法
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// 需要看最长递增子序列！
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// 尤其是理解ends数组的意义！
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public static int number2(int n, int m) {
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int[][][][] dp = new int[n][m + 1][m + 1][m + 1];
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for (int i = 0; i < n; i++) {
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for (int f = 0; f <= m; f++) {
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for (int s = 0; s <= m; s++) {
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for (int t = 0; t <= m; t++) {
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dp[i][f][s][t] = -1;
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}
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}
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}
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}
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return process2(0, 0, 0, 0, n, m, dp);
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}
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public static int process2(int i, int f, int s, int t, int n, int m, int[][][][] dp) {
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if (i == n) {
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return f != 0 && s != 0 && t != 0 ? 1 : 0;
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}
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if (dp[i][f][s][t] != -1) {
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return dp[i][f][s][t];
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}
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int ans = 0;
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for (int cur = 1; cur <= m; cur++) {
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if (f == 0 || cur <= f) {
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ans += process2(i + 1, cur, s, t, n, m, dp);
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} else if (s == 0 || cur <= s) {
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ans += process2(i + 1, f, cur, t, n, m, dp);
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} else if (t == 0 || cur <= t) {
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ans += process2(i + 1, f, s, cur, n, m, dp);
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}
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}
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dp[i][f][s][t] = ans;
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return ans;
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}
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public static void main(String[] args) {
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System.out.println("功能测试开始");
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for (int n = 4; n <= 8; n++) {
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for (int m = 1; m <= 5; m++) {
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int ans1 = number1(n, m);
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int ans2 = number2(n, m);
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if (ans1 != ans2) {
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System.out.println(ans1);
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System.out.println(ans2);
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System.out.println("出错了!");
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}
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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 = 2000;
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int m = 20;
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System.out.println("序列长度 : " + n);
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System.out.println("数字范围 : " + m);
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long start = System.currentTimeMillis();
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number2(n, m);
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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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