【Codeforces Round 365 (Div 2)C】【贪心 极限阻碍思想】 Chris and Road 过马路不被凸包车碰撞的最小时间

C. Chris and Road

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

And while Mishka is enjoying her trip...
Chris is a little brown bear. No one knows, where and when he met Mishka, but for a long time they are together (excluding her current trip). However, best friends are important
too. John is Chris' best friend.
Once walking with his friend, John gave Chris the following problem:
At the infinite horizontal road of width w,
bounded by lines y = 0 and y = w,
there is a bus moving, presented as a convex polygon ofn vertices. The bus moves continuously
with a constant speed of v in a straight Ox line
in direction of decreasing x coordinates, thus in time only x coordinates of
its points are changing. Formally, after time t each of x coordinates
of its points will be decreased byvt.
There is a pedestrian in the point (0, 0), who
can move only by a vertical pedestrian crossing, presented as a segment connecting points (0, 0) and (0, w) with
any speed not exceeding u. Thus the pedestrian can move only in a straight line Oy in
any direction with any speed not exceeding u and not leaving the road borders.
The pedestrian can instantly change his speed, thus, for example, he can stop instantly.
Please look at the sample note picture for better understanding.
We consider the pedestrian is hit by the bus, if at any moment the point he is located in lies strictly
inside the bus polygon (this means that if the point lies on the polygon vertex or on its edge, the pedestrian is not hit by the bus).
You are given the bus position at the moment 0.
Please help Chris determine minimum amount of time the pedestrian needs to cross the road and reach the point (0, w) and
not to be hit by the bus.

Input
The first line of the input contains four integers n, w, v, u (3 ≤ n ≤ 10 000, 1 ≤ w ≤ 109, 1 ≤ v,  u ≤ 1000) —
the number of the bus polygon vertices, road width, bus speed and pedestrian speed respectively.
The next n lines
describes polygon vertices in counter-clockwise order. i-th of them contains pair of integers xi and yi( - 109 ≤ xi ≤ 109, 0 ≤ yi ≤ w) —
coordinates of i-th polygon point. It is guaranteed that the polygon is non-degenerate.

Output
Print the single real t —
the time the pedestrian needs to croos the road and not to be hit by the bus. The answer is considered correct if its relative or absolute error doesn't exceed 10 - 6.

Example

input

5 5 1 2
1 2
3 1
4 3
3 4
1 4

output

5.0000000000

Note
Following image describes initial position in the first sample case:



#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
int n, w;
double u, v;
double x, y;
int main()
{
while (~scanf("%d%d%lf%lf", &n, &w, &v, &u))
{
bool early = 1;
bool late = 1;
double ans = 0;
for (int i = 1; i <= n; ++i)
{
scanf("%lf%lf", &x, &y);
if (x / v > y / u)early = 0;
if (x / v < y / u)late = 0;
gmax(ans, x / v + (w - y) / u);
}
if (early || late)ans = w / u;
printf("%.10f\n", ans);
}
return 0;
}
/*
【trick&&吐槽】
这道题看似是计算几何，实际其实是一个贪心问题。

【题意】
给你一个无线长的路，路的横坐标范围为[-∞,∞]
路的纵坐标范围为[0,w]
车速为v（稳定），人速为u（上限）
告诉你车子凸包每个点的坐标，
车从右往左开，人想从[0,0]到达[0,w]，只能向上走。
问你人过马路不被车撞的最小时间。

【类型】
贪心

【分析】
这道题看似是计算几何，实际其实是一个贪心问题。
首先，如果没有该车的话，这个人达到(0,w)的最小时间显然是w/u.
然后，车的每个点到达OY轴都有一个时间x/v，同时到达一个相应位置y
如果这个人到达任意一个y的时间y/u比车都要早的话，那直接走过了
如果这个人到达任意一个y的时间y/u比车都要晚的话，那直接走过了
如果不是以上两种情况，说明这个人会被车阻碍。
那显然，max((w-y)/v + x/v)就是答案

【时间复杂度&&优化】
O(n)

*/