Vacation（思维）

题目描述
Tom and Jerry are going on a vacation. They are now driving on a one-way road and several cars are in front of them. To be more specific, there are n cars in front of them. The ith car has a length of li, the head of it is si from the stop-line, and its maximum velocity is vi. The car Tom and Jerry are driving is l0 in length, and s0 from the stop-line, with a maximum velocity of v0.
The traffic light has a very long cycle. You can assume that it is always green light. However, since the road is too narrow, no car can get ahead of other cars. Even if your speed can be greater than the car in front of you, you still can only drive at the same speed as the anterior car. But when not affected by the car ahead, the driver will drive at the maximum speed. You can assume that every driver here is very good at driving, so that the distance of adjacent cars can be kept to be 0.
Though Tom and Jerry know that they can pass the stop-line during green light, they still want to know the minimum time they need to pass the stop-line. We say a car passes the stop-line once the head of the car passes it.
Please notice that even after a car passes the stop-line, it still runs on the road, and cannot be overtaken.

输入
This problem contains multiple test cases.
For each test case, the first line contains an integer n (1≤n≤105,∑n≤2×106), the number of cars.
The next three lines each contains n+1 integers, li,si,vi (1≤si,vi,li≤109). It’s guaranteed that si≥si+1+li+1,∀i∈[0,n−1]

输出
For each test case, output one line containing the answer. Your answer will be accepted if its absolute or relative error does not exceed 10−6.
Formally, let your answer be a, and the jury’s answer is b. Your answer is considered correct if |a−b|/max(1,|b|)≤10−6.
The answer is guaranteed to exist.

样例输入
1
2 2
7 1
2 1
2
1 2 2
10 7 1
6 2 1

样例输出
3.5000000000
5.0000000000

思路
自1到n递推每一辆车，计算该车与本车之间的相对速度
当该车速度大于等于本车时，则本车追不上该车，只计算该车阻拦时，所用的时间即为s0/v0
当该车速度小于本车时，当本车追不上该车，则只计算该车阻拦时的所用的时间为s0/v0，若能追上，则时间为追上的时间+以该车速度到达终点的时间

代码实现

#include<bits/stdc++.h>
using namespace std;

const int N=100005;
const int mod=998244353;
typedef long long ll;
typedef pair<int,int> P;
double l[N],s[N],v[N],sl[N];
int n;

int main()
{
    while(~scanf("%d",&n))
    {
        double ans=0;
        for(int i=0;i<=n;i++)
        {
            scanf("%lf",&l[i]);
            if(i) sl[i]=sl[i-1]+l[i];
        }
        for(int i=0;i<=n;i++) scanf("%lf",&s[i]);
        for(int i=0;i<=n;i++) scanf("%lf",&v[i]);
        for(int i=1;i<=n;i++)
        {
            double t;
            if(v[i]>=v[0]) t=s[0]/v[0];
            else
            {
                double tt=(s[0]-s[i]-sl[i])/(v[0]-v[i]);
                if(tt*v[0]>=s[0]) t=s[0]/v[0];
                else t=tt+(s[0]-tt*v[0])/v[i];
            }
            ans=max(ans,t);
        }
        printf("%.10f\n",ans);
    }
    return 0;
}