AtCoder Express(数学+二分)

D - AtCoder Express

Time limit : 2sec / Memory limit : 256MB

Score : 400 points

Problem Statement

In the year 2168, AtCoder Inc., which is much larger than now, is starting a limited express train service called AtCoder Express.

In the plan developed by the president Takahashi, the trains will run as follows:

  • A train will run for (t1+t2+t3+…+tN) seconds.
  • In the first t1 seconds, a train must run at a speed of at most v1 m/s (meters per second). Similarly, in the subsequent t2 seconds, a train must run at a speed of at mostv2 m/s, and so on.

According to the specifications of the trains, the acceleration of a train must be always within ±1m⁄s2. Additionally, a train must stop at the beginning and the end of the run.

Find the maximum possible distance that a train can cover in the run.

Constraints

  • 1≤N≤100
  • 1≤ti≤200
  • 1≤vi≤100
  • All input values are integers.

Input

Input is given from Standard Input in the following format:

N
t1 t2 t3 … tN
v1 v2 v3 … vN

Output

Print the maximum possible that a train can cover in the run.
Output is considered correct if its absolute difference from the judge's output is at most 10−3.

Sample Input 1

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1
100
30

Sample Output 1

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2100.000000000000000

The maximum distance is achieved when a train runs as follows:

  • In the first 30 seconds, it accelerates at a rate of 1m⁄s2, covering 450 meters.
  • In the subsequent 40 seconds, it maintains the velocity of 30m⁄s, covering 1200 meters.
  • In the last 30 seconds, it decelerates at the acceleration of −1m⁄s2, covering 450 meters.

The total distance covered is 450 + 1200 + 450 = 2100 meters.

Sample Input 2

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2
60 50
34 38

Sample Output 2

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2632.000000000000000

The maximum distance is achieved when a train runs as follows:

  • In the first 34 seconds, it accelerates at a rate of 1m⁄s2, covering 578 meters.
  • In the subsequent 26 seconds, it maintains the velocity of 34m⁄s, covering 884 meters.
  • In the subsequent 4 seconds, it accelerates at a rate of 1m⁄s2, covering 144 meters.
  • In the subsequent 8 seconds, it maintains the velocity of 38m⁄s, covering 304 meters.
  • In the last 38 seconds, it decelerates at the acceleration of −1m⁄s2, covering 722 meters.

The total distance covered is 578 + 884 + 144 + 304 + 722 = 2632 meters.

Sample Input 3

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3
12 14 2
6 2 7

Sample Output 3

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76.000000000000000

The maximum distance is achieved when a train runs as follows:

  • In the first 6 seconds, it accelerates at a rate of 1m⁄s2, covering 18 meters.
  • In the subsequent 2 seconds, it maintains the velocity of 6m⁄s, covering 12 meters.
  • In the subsequent 4 seconds, it decelerates at the acceleration of −1m⁄s2, covering 16 meters.
  • In the subsequent 14 seconds, it maintains the velocity of 2m⁄s, covering 28 meters.
  • In the last 2 seconds, it decelerates at the acceleration of −1m⁄s2, covering 2 meters.

The total distance covered is 18 + 12 + 16 + 28 + 2 = 76 meters.

Sample Input 4

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1
9
10

Sample Output 4

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20.250000000000000000

The maximum distance is achieved when a train runs as follows:

  • In the first 4.5 seconds, it accelerates at a rate of 1m⁄s2, covering 10.125 meters.
  • In the last 4.5 seconds, it decelerates at the acceleration of −1m⁄s2, covering 10.125 meters.

The total distance covered is 10.125 + 10.125 = 20.25 meters.

Sample Input 5

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10
64 55 27 35 76 119 7 18 49 100
29 19 31 39 27 48 41 87 55 70

Sample Output 5

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20291.000000000000




//题意：读了老半天，就是说一列火车在 n 段路上行驶，给出行驶的时间，和最大速度限制，起点终点速度要为 0 ，问最大可行驶的路程是多少?
//应该是头次做这种题吧， 所以，做的比较慢，先逆序扫一遍，可得到在每段路上行驶，为了到终点为 0 速度，限制速度是多少。
然后正序累加，具体操作是，二分（加速时间+匀速时间)的最大值，然后就可以轻松解决了
 1 # include <bits/stdc++.h>
 2 using namespace std;
 3 # define eps 1e-8
 4 # define INF 1e20
 5 # define pi  acos(-1.0)
 6 # define MX  105
 7 struct Node
 8 {
 9     double t,v;
10     double ev; //从终点的最大速度
11 }seg[MX];
12  
13 int n;
14 double spd, ans;
15  
16 void gogo(int x)
17 {
18     double l=0, r=seg[x].t;
19     double res = 0;
20     while (l<r-eps) // <
21     {
22         double mid = (l+r)/2;
23         if (mid + spd > seg[x].v-eps)
24         {
25             if ((seg[x].v - seg[x+1].ev) + mid < seg[x].t+eps)
26             {
27                 res = mid;
28                 l = mid;
29             }
30             else r = mid;
31         }
32         else
33         {
34             if (mid + spd - (seg[x].t-mid) - seg[x+1].ev < -eps) //<=
35             {
36                 res = mid;
37                 l = mid;
38             }
39             else r = mid;
40         }
41     }
42  
43     double jia, yun, jian;
44     if (spd + res < seg[x].v)
45     {
46         jia = res;
47         yun = 0;
48         jian = seg[x].t - jia;
49     }
50     else
51     {
52         jia = seg[x].v - spd;
53         yun = res - jia;
54         jian = seg[x].t - jia - yun;
55     }
56     ans += 1.0/2.0*jia*jia + spd * jia;
57     ans += yun * seg[x].v;
58  
59     if (res+spd < seg[x].v)
60     {
61         ans += -1.0/2.0*jian*jian +  (spd+jia) * jian;
62         spd = spd + jia - jian;
63     }
64     else
65     {
66         ans += -1.0/2.0*jian*jian +  seg[x].v * jian;
67         spd = seg[x].v - jian;
68     }
69  
70 }
71  
72 int main()
73 {
74     scanf("%d",&n);
75     for (int i=1;i<=n;i++)
76         scanf("%lf",&seg[i].t);
77     for (int i=1;i<=n;i++)
78         scanf("%lf",&seg[i].v);
79     spd=0.0;
80     for (int i=n;i>=1;i--)
81     {
82         if (spd+seg[i].t < seg[i].v+eps) //<=
83             spd += seg[i].t;
84         else
85             spd = seg[i].v;
86         seg[i].ev = spd;
87     }
88     ans = 0;
89     spd = 0;
90     seg[n+1] = (Node){0.0, 0.0, 0.0};
91  
92     for (int i=1;i<=n;i++)
93     {
94         gogo(i);
95     }
96     printf("%.5f\n",ans);
97     return 0;
98 }
View Code

 




posted @ 2017-10-29 19:52  happy_codes  阅读(532)  评论(0)  编辑  收藏  举报