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‎src/class086/Code01_LCS.java‎

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package class086;
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// 最长公共子序列其中一个结果
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// 给定两个字符串str1和str2
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// 输出两个字符串的最长公共子序列
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// 如果最长公共子序列为空，则输出-1
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// 测试链接 : https://www.nowcoder.com/practice/4727c06b9ee9446cab2e859b4bb86bb8
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// 请同学们务必参考如下代码中关于输入、输出的处理
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// 这是输入输出处理效率很高的写法
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// 提交以下的code，提交时请把类名改成"Main"，可以直接通过
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import java.io.BufferedReader;
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import java.io.IOException;
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import java.io.InputStreamReader;
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import java.io.OutputStreamWriter;
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import java.io.PrintWriter;
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// 讲解067 - 题目3，最长公共子序列长度
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public class Code01_LCS {
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public static int MAXN = 5001;
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public static int[][] dp = new int[MAXN][MAXN];
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public static char[] ans = new char[MAXN];
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public static char[] s1;
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public static char[] s2;
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public static int n, m, k;
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public static void main(String[] args) throws IOException {
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BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
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PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
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s1 = br.readLine().toCharArray();
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s2 = br.readLine().toCharArray();
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n = s1.length;
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m = s2.length;
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lcs();
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if (k == 0) {
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out.println(-1);
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} else {
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for (int i = 0; i < k; i++) {
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out.print(ans[i]);
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}
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out.println();
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}
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out.flush();
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out.close();
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br.close();
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}
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public static void lcs() {
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dp();
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k = dp[n][m];
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if (k > 0) {
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for (int len = k, i = n, j = m; len > 0;) {
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if (s1[i - 1] == s2[j - 1]) {
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ans[--len] = s1[i - 1];
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i--;
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j--;
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} else {
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if (dp[i - 1][j] >= dp[i][j - 1]) {
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i--;
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} else {
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j--;
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}
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}
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}
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}
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}
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// 填好dp表
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public static void dp() {
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for (int i = 1; i <= n; i++) {
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for (int j = 1; j <= m; j++) {
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if (s1[i - 1] == s2[j - 1]) {
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dp[i][j] = 1 + dp[i - 1][j - 1];
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} else {
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dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
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}
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}
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}
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}
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}

‎src/class086/Code02_LIS.java‎

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package class086;
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// 最长递增子序列字典序最小的结果
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// 给定数组arr，设长度为n
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// 输出arr的最长递增子序列
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// 如果有多个答案，请输出其中字典序最小的
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// 字典序设定 :
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// 每个数字看作是单独的字符，比如120认为比36的字典序大
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// 测试链接 : https://www.nowcoder.com/practice/30fb9b3cab9742ecae9acda1c75bf927
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// 请同学们务必参考如下代码中关于输入、输出的处理
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// 这是输入输出处理效率很高的写法
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// 提交以下的code，提交时请把类名改成"Main"，可以直接通过
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import java.io.BufferedReader;
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import java.io.IOException;
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import java.io.InputStreamReader;
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import java.io.OutputStreamWriter;
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import java.io.PrintWriter;
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import java.io.StreamTokenizer;
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import java.util.Arrays;
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// 讲解072 - 最长递增子序列及其扩展
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public class Code02_LIS {
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public static int MAXN = 100001;
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public static int[] nums = new int[MAXN];
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public static int[] ends = new int[MAXN];
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public static int[] dp = new int[MAXN];
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public static int[] ans = new int[MAXN];
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public static int n, k;
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public static void main(String[] args) throws IOException {
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BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
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StreamTokenizer in = new StreamTokenizer(br);
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PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
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while (in.nextToken() != StreamTokenizer.TT_EOF) {
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n = (int) in.nval;
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for (int i = 0; i < n; i++) {
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in.nextToken();
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nums[i] = (int) in.nval;
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}
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lis();
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for (int i = 0; i < k - 1; i++) {
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out.print(ans[i] + " ");
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}
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out.println(ans[k - 1]);
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}
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out.flush();
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out.close();
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br.close();
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}
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public static void lis() {
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k = dp();
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Arrays.fill(ans, 0, k, Integer.MAX_VALUE);
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for (int i = 0; i < n; i++) {
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if (dp[i] == k) {
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if (nums[i] < ans[0]) {
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ans[0] = nums[i];
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}
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} else {
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if (ans[k - dp[i] - 1] < nums[i] && nums[i] < ans[k - dp[i]]) {
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ans[k - dp[i]] = nums[i];
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}
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}
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}
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}
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// dp[i] : 必须以i位置的数字开头的情况下，最长递增子序列长度
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// 填好dp表 + 返回最长递增子序列长度
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public static int dp() {
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int len = 0;
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for (int i = n - 1, find; i >= 0; i--) {
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find = bs(len, nums[i]);
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if (find == -1) {
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ends[len++] = nums[i];
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dp[i] = len;
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} else {
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ends[find] = nums[i];
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dp[i] = find + 1;
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}
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}
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return len;
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}
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public static int bs(int len, int num) {
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int l = 0, r = len - 1, m, ans = -1;
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while (l <= r) {
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m = (l + r) / 2;
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if (ends[m] <= num) {
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ans = m;
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r = m - 1;
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} else {
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l = m + 1;
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}
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}
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return ans;
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}
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}

‎src/class086/Code03_Diving.java‎

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package class086;
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// 潜水的最大时间与方案
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// 一共有n个工具，每个工具都有自己的重量a、阻力b、提升的停留时间c
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// 因为背包有限，所以只能背重量不超过m的工具
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// 因为力气有限，所以只能背阻力不超过v的工具
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// 希望能在水下停留的时间最久
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// 返回最久的停留时间和下标字典序最小的选择工具的方案
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// 字典序设定 :
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// 下标的数值认为是字符串，比如下标120认为比下标36的字典序小
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// 测试链接 : https://www.luogu.com.cn/problem/P1759
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// 请同学们务必参考如下代码中关于输入、输出的处理
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// 这是输入输出处理效率很高的写法
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// 提交以下的code，提交时请把类名改成"Main"，可以直接通过
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import java.io.BufferedReader;
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import java.io.IOException;
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import java.io.InputStreamReader;
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import java.io.OutputStreamWriter;
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import java.io.PrintWriter;
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import java.io.StreamTokenizer;
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// 讲解069 - 多维费用背包 + 空间压缩
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public class Code03_Diving {
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public static int MAXN = 101;
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public static int MAXM = 201;
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public static int[] a = new int[MAXN];
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public static int[] b = new int[MAXN];
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public static int[] c = new int[MAXN];
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public static int[][] dp = new int[MAXM][MAXM];
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public static String[][] path = new String[MAXM][MAXM];
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public static int m, v, n;
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public static void build() {
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for (int i = 0; i <= m; i++) {
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for (int j = 0; j <= v; j++) {
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dp[i][j] = 0;
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path[i][j] = null;
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}
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}
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}
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public static void main(String[] args) throws IOException {
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BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
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StreamTokenizer in = new StreamTokenizer(br);
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PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
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while (in.nextToken() != StreamTokenizer.TT_EOF) {
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m = (int) in.nval;
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in.nextToken();
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v = (int) in.nval;
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in.nextToken();
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n = (int) in.nval;
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build();
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for (int i = 1; i <= n; i++) {
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in.nextToken();
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a[i] = (int) in.nval;
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in.nextToken();
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b[i] = (int) in.nval;
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in.nextToken();
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c[i] = (int) in.nval;
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}
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compute();
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out.println(dp[m][v]);
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out.println(path[m][v]);
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}
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out.flush();
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out.close();
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br.close();
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}
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// 这个方法就是讲了很多遍的空间压缩技巧
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// 请务必掌握动态规划前几节课对这一技巧的讲解
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public static void compute() {
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String cur;
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for (int i = 1; i <= n; i++) {
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for (int j = m; j >= a[i]; j--) {
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for (int k = v; k >= b[i]; k--) {
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if (path[j - a[i]][k - b[i]] == null) {
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cur = String.valueOf(i);
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} else {
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cur = path[j - a[i]][k - b[i]] + " " + String.valueOf(i);
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}
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if (dp[j][k] < dp[j - a[i]][k - b[i]] + c[i]) {
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dp[j][k] = dp[j - a[i]][k - b[i]] + c[i];
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path[j][k] = cur;
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} else if (dp[j][k] == dp[j - a[i]][k - b[i]] + c[i]) {
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if (cur.compareTo(path[j][k]) < 0) {
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path[j][k] = cur;
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}
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}
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}
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}
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}
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}
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}

‎src/class086/Code04_SmallestSufficientTeam.java‎

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package class086;
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import java.util.Arrays;
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import java.util.List;
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// 最小的必要团队
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// 作为项目经理，你规划了一份需求的技能清单req_skills
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// 并打算从备选人员名单people中选出些人组成必要团队
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// 编号为i的备选人员people[i]含有一份该备选人员掌握的技能列表
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// 所谓必要团队，就是在这个团队中
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// 对于所需求的技能列表req_skills中列出的每项技能，团队中至少有一名成员已经掌握
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// 请你返回规模最小的必要团队，团队成员用人员编号表示
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// 你可以按 任意顺序 返回答案，题目数据保证答案存在
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// 测试链接 : https://leetcode.cn/problems/smallest-sufficient-team/
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public class Code04_SmallestSufficientTeam {
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// 讲解080、081 - 状压dp
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public static int[] smallestSufficientTeam(String[] skills, List<List<String>> people) {
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Arrays.sort(skills);
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int n = skills.length;
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int m = people.size();
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int[] arr = new int[m];
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for (int i = 0; i < m; i++) {
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int status = 0;
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List<String> skill = people.get(i);
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skill.sort((a, b) -> a.compareTo(b));
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int p1 = 0;
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int p2 = 0;
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while (p1 < n && p2 < skill.size()) {
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int compare = skills[p1].compareTo(skill.get(p2));
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if (compare < 0) {
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p1++;
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} else if (compare > 0) {
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p2++;
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} else {
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status |= 1 << p1;
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p1++;
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p2++;
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}
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}
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arr[i] = status;
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}
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int[][] dp = new int[m][1 << n];
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for (int i = 0; i < m; i++) {
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Arrays.fill(dp[i], -1);
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}
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int size = f(arr, n, 0, 0, dp);
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int[] ans = new int[size];
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int ansi = 0;
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int i = 0;
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int status = 0;
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while (status != (1 << n) - 1) {
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if (i + 1 == m || dp[i][status] != dp[i + 1][status]) {
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ans[ansi++] = i;
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status |= arr[i];
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}
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i++;
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}
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return ans;
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}
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public static int f(int[] arr, int n, int i, int status, int[][] dp) {
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if (status == (1 << n) - 1) {
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return 0;
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}
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if (i == arr.length) {
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return Integer.MAX_VALUE;
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}
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if (dp[i][status] != -1) {
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return dp[i][status];
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}
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int p1 = f(arr, n, i + 1, status, dp);
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int p2 = Integer.MAX_VALUE;
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int next2 = f(arr, n, i + 1, status | arr[i], dp);
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if (next2 != Integer.MAX_VALUE) {
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p2 = 1 + next2;
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}
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int ans = Math.min(p1, p2);
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dp[i][status] = ans;
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return ans;
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}
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}

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