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Crawling in process... Crawling failed Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
Description
Input
Output
Sample Input
Sample Output
Hint
Description
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 of n 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 by vt.
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.
Sample Input
5 5 1 2
1 2
3 1
4 3
3 4
1 4
5.0000000000
Sample Output
Hint
Following image describes initial position in the first sample case:
/* 赛后除了F题才想这个B题的，被A题坑死了 先判断一下是否能在车来之前就过去，不能的话就按着最右边的点走 第二种情况很简单，只需要先按照y坐标排序，按照时间的顺序走x就行了 */ # include <iostream> # include <stdio.h> # include <vector> # include <algorithm> # include <cstdio> using namespace std; struct Point { double x; double y; }; bool cmp(Point a, Point b) { if(a.y != b.y) return a.y < b.y; return a.x < b.x; } int main() { double n, w, v, u; while(scanf("%lf%lf%lf%lf",&n,&w,&v,&u)!=EOF) { vector <Point> m_v; for(int i = 0; i < n; i++) { Point p; scanf("%lf%lf",&p.x,&p.y); m_v.push_back(p); } sort(m_v.begin(), m_v.end(), cmp); bool ok1 = 1, ok2 = 1; double t = 0; for(int i = 0; i < m_v.size(); i++) { if (m_v[i].x / v > m_v[i].y / u) ok1 = 0; if (m_v[i].x / v < m_v[i].y / u) ok2 = 0; t = max(t, m_v[i].x / v + (w - m_v[i].y) / u); } if(ok1 || ok2) printf("%.10lf\n", w / u); else printf("%.10lf\n", t); } return 0; }
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原文地址：http://www.cnblogs.com/wuwangchuxin0924/p/5754693.html
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