思路: 签到题

按题意描述编写

import java.io.*;
import java.util.*;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(new BufferedInputStream(System.in));int a = sc.nextInt(), b = sc.nextInt(), c = sc.nextInt(), d = sc.nextInt();System.out.println((a + b) < (c + d) ? "YES" : "NO");}}

思路: 偏序

经典的俄罗斯套娃模型，典题

主要通过 排序 做文章

• 一级排序，左端点从小到大
• 二级排序，右端点从大到小

然后维护最大的右端点区间id即可，不需要treemap维护了

时间复杂度为$O(nlogn)$

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;public class Main {public static void main(String[] args) {AReader sc = new AReader();int n = sc.nextInt();int[][] intervals = new int[n][3];for (int i = 0; i < n; i++) {intervals[i][0] = sc.nextInt();intervals[i][1] = sc.nextInt();intervals[i][2] = i;}Arrays.sort(intervals, (a, b) -> {if (a[0] != b[0]) return a[0] < b[0] ? -1 : 1;if (a[1] != b[1]) return a[1] > b[1] ? -1 : 1;return 0;});int ansI = -1, ansJ = -1;// 偏序关系吗？TreeMap<Integer, Integer> tree = new TreeMap<>();for (int i = 0; i < n; i++) {int right = intervals[i][1];Map.Entry<Integer, Integer> ent = tree.ceilingEntry(right);if (ent != null) {ansI = intervals[i][2] + 1;ansJ = ent.getValue() + 1;break;}tree.put(right, intervals[i][2]);}System.out.println(ansI + " " + ansJ);}staticclass AReader {private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));private StringTokenizer tokenizer = new StringTokenizer("");private String innerNextLine() {try {return reader.readLine();} catch (IOException ex) {return null;}}public boolean hasNext() {while (!tokenizer.hasMoreTokens()) {String nextLine = innerNextLine();if (nextLine == null) {return false;}tokenizer = new StringTokenizer(nextLine);}return true;}public String nextLine() {tokenizer = new StringTokenizer("");return innerNextLine();}public String next() {hasNext();return tokenizer.nextToken();}public int nextInt() {return Integer.parseInt(next());}public long nextLong() {return Long.parseLong(next());}//        public BigInteger nextBigInt() {
//            return new BigInteger(next());
//        }// 若需要nextDouble等方法，请自行调用Double.parseDouble包装}}

思路: DP + 回溯数组

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.util.stream.Collectors;public class Main {public static void main(String[] args) {AReader sc = new AReader();int n = sc.nextInt();int[] arr = new int[n];for (int i = 0; i < n; i++) {arr[i] = sc.nextInt();}Map<Integer, Integer> hash = new HashMap<>();int[] dp = new int[n];int[] from = new int[n];Arrays.fill(from,  -1);for (int i = 0; i < n; i++) {int v = arr[i];if (hash.containsKey(v - 1)) {dp[i] = dp[hash.get(v - 1)] + 1;from[i] = hash.get(v - 1);} else {dp[i] = 1;from[i] = -1;}hash.put(v, i);}int idx = -1;for (int i = 0; i < n; i++) {if (idx == -1 || dp[idx] < dp[i]) {idx = i;}}System.out.println(dp[idx]);List<Integer> lists = new ArrayList<>();while (idx != -1) {lists.add(idx + 1);idx = from[idx];}Collections.reverse(lists);System.out.println(lists.stream().map(String::valueOf).collect(Collectors.joining(" ")));}staticclass AReader {private BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));private StringTokenizer tokenizer = new StringTokenizer("");private String innerNextLine() {try {return reader.readLine();} catch (IOException ex) {return null;}}public boolean hasNext() {while (!tokenizer.hasMoreTokens()) {String nextLine = innerNextLine();if (nextLine == null) {return false;}tokenizer = new StringTokenizer(nextLine);}return true;}public String nextLine() {tokenizer = new StringTokenizer("");return innerNextLine();}public String next() {hasNext();return tokenizer.nextToken();}public int nextInt() {return Integer.parseInt(next());}public long nextLong() {return Long.parseLong(next());}//        public BigInteger nextBigInt() {
//            return new BigInteger(next());
//        }// 若需要nextDouble等方法，请自行调用Double.parseDouble包装}}