AtCoder Express(数学+二分)

D - AtCoder Express

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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.

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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.

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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.

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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.

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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.

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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.

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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

# include <bits/stdc++.h>
using namespace std;
# define eps 1e-8
# define INF 1e20
# define pi  acos(-1.0)
# define MX  105
struct Node
{
    double t,v;
    double ev; //从终点的最大速度
}seg[MX];
 
int n;
double spd, ans;
 
void gogo(int x)
{
    double l=0, r=seg[x].t;
    double res = 0;
    while (l<r-eps) // <
    {
        double mid = (l+r)/2;
        if (mid + spd > seg[x].v-eps)
        {
            if ((seg[x].v - seg[x+1].ev) + mid < seg[x].t+eps)
            {
                res = mid;
                l = mid;
            }
            else r = mid;
        }
        else
        {
            if (mid + spd - (seg[x].t-mid) - seg[x+1].ev < -eps) //<=
            {
                res = mid;
                l = mid;
            }
            else r = mid;
        }
    }
 
    double jia, yun, jian;
    if (spd + res < seg[x].v)
    {
        jia = res;
        yun = 0;
        jian = seg[x].t - jia;
    }
    else
    {
        jia = seg[x].v - spd;
        yun = res - jia;
        jian = seg[x].t - jia - yun;
    }
    ans += 1.0/2.0*jia*jia + spd * jia;
    ans += yun * seg[x].v;
 
    if (res+spd < seg[x].v)
    {
        ans += -1.0/2.0*jian*jian +  (spd+jia) * jian;
        spd = spd + jia - jian;
    }
    else
    {
        ans += -1.0/2.0*jian*jian +  seg[x].v * jian;
        spd = seg[x].v - jian;
    }
 
}
 
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
        scanf("%lf",&seg[i].t);
    for (int i=1;i<=n;i++)
        scanf("%lf",&seg[i].v);
    spd=0.0;
    for (int i=n;i>=1;i--)
    {
        if (spd+seg[i].t < seg[i].v+eps) //<=
            spd += seg[i].t;
        else
            spd = seg[i].v;
        seg[i].ev = spd;
    }
    ans = 0;
    spd = 0;
    seg[n+1] = (Node){0.0, 0.0, 0.0};
 
    for (int i=1;i<=n;i++)
    {
        gogo(i);
    }
    printf("%.5f
",ans);
    return 0;
}