hdu 6851 Vacation (thinking + greedy)

Portal

• the meaning of problems

Numbered 0 to n, n + 1 line have traffic lights vehicle, from 0 to n from the line closer traffic lights

Each will have a maximum speed v, body length l, and the distance from the line of traffic lights s,

A front reaches the line then the car has reached the line

If a car is not close to the front of the car, so the car can travel at a maximum speed

If the front next to a car, close to the front before a rear travel, can not overtake!

Even after the traffic light line can not overtake!

Asked the shortest time of the first 0 is offline farthest one, to reach the line

• ideas

Since I can not overtake, then there are two possibilities when the furthest car to the line

① own line to $ t = \ frac {s_ {0}} {v_ {0}} $

② connected to the car after the first line p $ \ frac {s_ {p} + \ sum_ {1} ^ {p} l_ {i}} {v_ {p}} $

Explain why this is so then counted in the back of a car

 Suppose there are two cars $ a_ {1} $ velocity $ v $, $ a_ {0} $ speed 2V $ $

When the current vehicle reaches $ p $ surface points, while the back of the car to reach $ {p} '$, both the time spent on the same

Similarly seen

When $ a_ {1} $ O $ $ arrival point, $ a_ {0} $ reach $ l_ {1} $ point, i.e. $ a_ {0} $ distance from the finish line is $ a_ {1} $ of Car length

② even if the formula is in the first car behind the $ p $ ($ p $ behind the cars are all connected up)

When the first car to reach the finish line $ p $, the farthest distance from the car to the finish line is that all $ p $ car length behind the car - a car last longer own car

I.e. $ sum = \ sum_ {1} ^ {p} l_ {i} $

Therefore traveling time is $ p $ + vehicle traveling time to the furthest car $ v_ {p} $ $ SUM with time i.e. $ $ \ frac {s_ {p}} {v_ {p}} + \ frac {\ sum_ {1} ^ {p} l_ {i}} {v_ {p}} $

Since the ground behind the car uncertain, it is possible to calculate the maximum time,

Travel time is long explanation slow speed behind will certainly pick behind him

• Code

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn=1e6+6;
 4 double l[maxn],s[maxn],v[maxn];
 5 double sum[maxn];
 6 int main()
 7 {
 8     int n;
 9     while(~scanf("%d",&n))
10     {
11         for(int i=0;i<=n;i++)
12         {
13             scanf("%lf",&l[i]);
14             if(i==0)
15                 continue;
16             sum[i]=sum[i-1]+l[i];
17         }
18         for(int i=0;i<=n;i++)
19             scanf("%lf",&s[i]);
20         for(int i=0;i<=n;i++)
21             scanf("%lf", & V [I]);
 22 is  
23 is          /// alone across the finish 
24          Double ANS = S [ 0 ] / V [ 0 ];
 25  
26 is          /// line through the end point 
27          for ( int I = . 1 ; I <= n- ; I ++ )
 28              ANS = max (ANS, (S [I] + SUM [I]) / V [I]);
 29  
30          the printf ( " % .10f \ n- " , ANS);
 31 is      }
 32 }

View Code