E. Hiking Dinkelbach

http://codeforces.com/contest/489/problem/E

 

 https://www.cnblogs.com/KirisameMarisa/p/4187637.html

 

 

 1 import java.util.Scanner;
 2 
 3 import static java.lang.Math.abs;
 4 import static java.lang.Math.sqrt;
 5 
 6 public class main2 {
 7     static Scanner io = new Scanner(System.in);
 8     static final int maxn = 1050, INF = (int) 1e9;
 9     static final double Eps = 1e-9;
10 
11     static int[] x = new int[maxn];
12     static int[] p = new int[maxn];
13 
14     static int[] pre = new int[maxn];
15 
16     static int n, L;
17 
18 
19     public static void main(String[] args) {
20         n = io.nextInt();
21         L = io.nextInt();
22         for (int i = 1; i <= n; i++) {
23             x[i] = io.nextInt();
24             p[i] = io.nextInt();
25         }
26 
27         //Dinkelbach
28         double ans = 1, tmp = 0;
29         double[] d = new double[maxn];
30         while (abs(ans - tmp) > Eps) {
31             ans = tmp;
32 
33             //博客的原话是“把最大的K个d[i]加起来就完事了”
34             //本题中只要d[i]>0就可以加起来，具备最优子结构，变成了简单的动态规划。而原题是不具备的
35 
36             for (int i = 1; i <= n; i++) {
37                 //本题求的是最小值，博客里是最大值，所以初始化要大，更新要小
38                 d[i] = INF;
39                 for (int j = 0; j < i; j++) {
40                     if (d[i] > d[j] + sqrt(abs(x[i] - x[j] - L)) - tmp * p[i]) {
41                         d[i] = d[j] + sqrt(abs(x[i] - x[j] - L)) - tmp * p[i];
42                         pre[i] = j;
43                     }
44                 }
45             }
46 
47             a=b=0;
48             add(n);
49             tmp = a/b;
50         }
51 
52         printt(n);
53     }
54 
55     static int a, b;
56 
57     static void add(int n) {
58         if (n == 0) return;
59         a += sqrt(abs(x[n] - x[pre[n]]-L));
60         b += p[n];
61         add(pre[n]);
62     }
63 
64     static void printt(int n){
65         if (n==0)return;
66         printt(pre[n]);
67         System.out.print(n+" ");
68     }
69 }