[测试题]line

Description

Input

Output

Sample Input

10 4974
3636 3
6679 70
7182 93
10618 98
14768 23
15242 99
16077 35
23368 46
27723 15
32404 81

Sample Output

0.465308

Hint

题解

典型的 $01$ 分数规划。

我们假设 $x[]$ 中和 $a[]$ 中只保存了选取的点。

依题，要求 $$ans={{sum_i {sqrt{|x[i]-x[i-1]-l|}}} over {sum_i a[i]}}$$

按照 $01$ 分数规划的套路，我们假设有对于当前值更优的解。

$$egin{aligned}ans&geq{{sum_i {sqrt{|x[i]-x[i-1]-l|}}} over {sum_i a[i]}}\ans imes{sum_i a[i]}&geq{sum_i {sqrt{|x[i]-x[i-1]-l|}}}\{sum_i ({ans imes a[i]})}&geq{sum_i {sqrt{|x[i]-x[i-1]-l|}}}\0&geq{sum_i {sqrt{|x[i]-x[i-1]-l|}}}-{sum_i ({ans imes a[i]})}\0&geq{sum_i ({sqrt{|x[i]-x[i-1]-l|}-ans imes a[i]})}end{aligned}$$

那么我们去二分这个 $ans$，每次做一次 $ ext{DP}$ 判断有无更优解即可。

//It is made by Awson on 2017.9.19
#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int INF = ~0u>>1;
const int N = 1000; 
const double eps = 1e-7;

int n, l;
int x[N+5], a[N+5];

bool judge(double r) {
    double f[N+5];
    f[0] = 0;
    for (int i = 1; i <= n; i++) {
        f[i] = INF;
        for (int j = 0; j < i; j++)
            f[i] = min(f[i], f[j]+sqrt(Abs(x[i]-x[j]-l))-r*a[i]);
    }
    return f[n] < eps;
}
void work() {
    for (int i = 1; i <= n; i++)
        scanf("%d%d", &x[i], &a[i]);
    double L = 0, R = 1e9;
    while (R - L > eps) {
        double mid = (L+R)/2.;
        if (judge(mid)) R = mid;
        else L = mid;
    }
    printf("%.6lf
", L);
}

int main() {
    freopen("line.in", "r", stdin);
    freopen("line.out", "w", stdout);
    while (~scanf("%d%d", &n, &l))
        work();
    return 0;
}