本文主要是介绍(AtCoder Beginner Contest 334) --- F - Christmas Present 2 -- 题解，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
F - Christmas Present 2
F - Christmas Present 2
题目大意： 
思路解析：
因为他是顺序前往每个孩子的家，前往时必须要带一个礼物，并且最多只能带k个礼物，所以它每次前往最多k个孩子之后就要回到初始点重新出发。
然后我们直接计算从初始点不回家顺序前往每个孩子的距离之和ans。再维护一个更新数组d[i]，更新从i回到初始点，再从初始点到i+1的额外开销。
再维护一个额外开销总和的数组， dp[i],表示从初始点走到 i个孩子的额外开销之和。
转移方程为 dp[i] = min(dp[j]) + d[i-1] Math.max(0,i-k)<=j<=i-1。min(dp[j])可以使用线段树维护实现。
这样我们就可以得到从初始点到第n-1个孩子再到初始点的额外开销之和，ans + dp[n]则为最终答案。
代码实现：
import java.io.*;
import java.math.BigInteger;
import java.util.*;public class Main {static int mod = 998244353;static double[] t = new double[200005];static double[] dp;static int n;public static void main(String[] args) throws IOException {int T = 1;for (int o = 0; o < T; o++) {n = input.nextInt();int k = input.nextInt();int X = input.nextInt();int Y = input.nextInt();int[][] xy = new int[n + 1][2];for (int i = 0; i < n; i++) {xy[i][0] = input.nextInt();xy[i][1] = input.nextInt();}xy[n][0] = X;xy[n][1] = Y;double ans = dist(X, Y, xy[0][0], xy[0][1]);double[] d = new double[n];for (int i = 0; i < n; i++) {double dir = dist(xy[i + 1][0], xy[i + 1][1], xy[i][0], xy[i][1]);double ret = dist(X, Y, xy[i][0], xy[i][1]) + dist(X, Y, xy[i + 1][0], xy[i + 1][1]);ans += dir;d[i] = ret - dir;}dp = new double[n + 1]; // dp[i][1][j] 表示当前已经给了i个孩子，并且在第i个位置，还能再送j个孩子Arrays.fill(dp, Double.MAX_VALUE);Arrays.fill(t, Double.MAX_VALUE);for (int i = 1; i <= n; i++) {double qre = query(Math.max(1, i - k), i-1);if (i - k <= 0) qre = Math.min(qre, 0);dp[i] = qre + d[i - 1];update(i, dp[i]);}ans += dp[n];out.printf("%.9f", ans);}out.flush();out.close();br.close();}public static int lowbit(int x) {return x & (-x);}public static void update(int i, double val) {while (i <= n) {t[i] = Math.min(t[i], val);i += lowbit(i);}}public static double query(int l, int r) {double res = Double.MAX_VALUE;while (r >= l) {res = Math.min(res, dp[r]);r--;while (r - l >= lowbit(r)) {res = Math.min(res, t[r]);r -= lowbit(r);}}return res;}public static double dist(int X, int Y, int x, int y) {return Math.sqrt((long) (X - x) * (X - x) + (long) (Y - y) * (Y - y));}static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));static Input input = new Input(System.in);static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));static class Input {public BufferedReader reader;public StringTokenizer tokenizer;public Input(InputStream stream) {reader = new BufferedReader(new InputStreamReader(stream), 32768);tokenizer = null;}public String next() {while (tokenizer == null || !tokenizer.hasMoreTokens()) {try {tokenizer = new StringTokenizer(reader.readLine());} catch (IOException e) {throw new RuntimeException(e);}}return tokenizer.nextToken();}public String nextLine() {String str = null;try {str = reader.readLine();} catch (IOException e) {// TODO 自动生成的 catch 块e.printStackTrace();}return str;}public int nextInt() {return Integer.parseInt(next());}public long nextLong() {return Long.parseLong(next());}public Double nextDouble() {return Double.parseDouble(next());}public BigInteger nextBigInteger() {return new BigInteger(next());}}}
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