Codeforces Gym 100531J Joy of Flight 变换坐标系
Joy of Flight
题目连接:
http://codeforces.com/gym/100531/attachments
Description
Jacob likes to play with his radio-controlled aircraft. The weather today is pretty windy and Jacob has
to plan flight carefully. He has a weather forecast — the speed and direction of the wind for every second
of the planned flight.
The plane may have airspeed up to vmax units per second in any direction. The wind blows away plane
in the following way: if airspeed speed of the plane is (vx, vy) and the wind speed is (wx, wy), the plane
moves by (vx + wx, vw + wy) each second.
Jacob has a fuel for exactly k seconds, and he wants to learn, whether the plane is able to fly from start
to finish in this time. If it is possible he needs to know the flight plan: the position of the plane after
every second of flight.
Input
The first line of the input file contains four integers Sx, Sy, Fx, Fy — coordinates of start and finish
(−10 000 ≤ Sx, Sy, Fx, Fy ≤ 10 000).
The second line contains three integers n, k and vmax — the number of wind condition changes, duration
of Jacob’s flight in seconds and maximum aircraft speed (1 ≤ n, k, vmax ≤ 10 000).
The following n lines contain the wind conditions description. The i-th of these lines contains integers
ti
Output
The first line must contain “Yes” if Jacob’s plane is able to fly from start to finish in k seconds, and “No”
otherwise.
If it can to do that, the following k lines must contain the flight plan. The i-th of these lines must contain
two floating point numbers x and y — the coordinates of the position (Pi) of the plane after i-th second
of the flight.
The plan is correct if for every 1 ≤ i ≤ k it is possible to fly in one second from Pi−1 to some point
Qi
, such that distance between Qi and Pi doesn’t exceed 10−5
, where P0 = S. Moreover the distance
between Pk and F should not exceed 10−5 as well.
Sample Input
1 1 7 4
2 3 10
0 1 2
2 2 0
Sample Output
Yes
3 2.5
5 2.5
7 4
Hint
题意
给你一个起点和终点,然后会有风,问你能否在k秒内到达终点
如果可以的话,就输出每一秒之后,你在哪儿
题解:
和CF的某道题一样的,把坐标系变换一下,把风直接的按在终点倒着跑
然后直接看一下时间是否小于等于k就好了
如果小于的话,就直接跑平均速度就好了
代码
#include<bits/stdc++.h>
using namespace std;
double sx,sy,fx,fy;
int n,k,t[20005];
double v,wx[20005],wy[20005];
double vx,vy;
int tot = 0;
double winx=0,winy=0;
int check(int mid)
{
tot = 0,winx = winy = 0;
double ex = fx,ey = fy;
for(int i=0;i<mid;i++)
{
if(tot<n&&t[tot]==i)
winx=wx[tot],winy=wy[tot],tot++;
ex-=winx,ey-=winy;
}
double dis = sqrt((ex-sx)*(ex-sx)+(ey-sy)*(ey-sy));
if(dis/v>mid)return 0;
return 1;
}
int main()
{
freopen("joy.in","r",stdin);
freopen("joy.out","w",stdout);
cin>>sx>>sy>>fx>>fy;
cin>>n>>k>>v;
for(int i=0;i<n;i++)
cin>>t[i]>>wx[i]>>wy[i];
int L = 0;
for(;L<=k+5;L++)
if(check(L))
break;
if(L>k)return puts("No");
double ex = fx,ey = fy;
tot = winx = winy = 0;
for(int i=0;i<L;i++)
{
if(tot<n&&t[tot]==i)
winx=wx[tot],winy=wy[tot],tot++;
fx-=winx,fy-=winy;
}
double l = fx - sx;
double h = fy - sy;
double dis = sqrt(l*l+h*h);
double vx = v * l / dis;
double vy = v * h / dis;
tot = winx = winy = 0;
puts("Yes");
if(L==0)L++;
for(int i=0;i<L-1;i++)
{
if(tot<n&&t[tot]==i)
winx=wx[tot],winy=wy[tot],tot++;
sx += vx + winx;
sy += vy + winy;
printf("%.6f %.6f\n",sx,sy);
}
for(int i=L;i<=k;i++)
printf("%.6f %.6f\n",ex,ey);
}
Codeforces Gym 100531J Joy of Flight 变换坐标系的更多相关文章
- Gym 100531J Joy of Flight (几何)
题意:你从开始坐标到末尾坐标,要经过 k 秒,然后给你每秒的风向,和飞机的最大速度,问能不能从开始到末尾. 析:首先这个风向是不确定的,所以我们先排除风向的影响,然后算出,静风是的最小速度,如果这都大 ...
- Codeforces Gym 101252D&&floyd判圈算法学习笔记
一句话题意:x0=1,xi+1=(Axi+xi%B)%C,如果x序列中存在最早的两个相同的元素,输出第二次出现的位置,若在2e7内无解则输出-1. 题解:都不到100天就AFO了才来学这floyd判圈 ...
- Codeforces Gym 101190M Mole Tunnels - 费用流
题目传送门 传送门 题目大意 $m$只鼹鼠有$n$个巢穴,$n - 1$条长度为$1$的通道将它们连通且第$i(i > 1)$个巢穴与第$\left\lfloor \frac{i}{2}\rig ...
- Codeforces Gym 101623A - 动态规划
题目传送门 传送门 题目大意 给定一个长度为$n$的序列,要求划分成最少的段数,然后将这些段排序使得新序列单调不减. 考虑将相邻的相等的数缩成一个数. 假设没有分成了$n$段,考虑最少能够减少多少划分 ...
- 【Codeforces Gym 100725K】Key Insertion
Codeforces Gym 100725K 题意:给定一个初始全0的序列,然后给\(n\)个查询,每一次调用\(Insert(L_i,i)\),其中\(Insert(L,K)\)表示在第L位插入K, ...
- Codeforces gym 101343 J.Husam and the Broken Present 2【状压dp】
2017 JUST Programming Contest 2.0 题目链接:Codeforces gym 101343 J.Husam and the Broken Present 2 J. Hu ...
- codeforces gym 100553I
codeforces gym 100553I solution 令a[i]表示位置i的船的编号 研究可以发现,应是从中间开始,往两边跳.... 于是就是一个点往两边的最长下降子序列之和减一 魔改树状数 ...
- CodeForces Gym 100213F Counterfeit Money
CodeForces Gym题目页面传送门 有\(1\)个\(n1\times m1\)的字符矩阵\(a\)和\(1\)个\(n2\times m2\)的字符矩阵\(b\),求\(a,b\)的最大公共 ...
- Codeforces GYM 100876 J - Buying roads 题解
Codeforces GYM 100876 J - Buying roads 题解 才不是因为有了图床来测试一下呢,哼( 题意 给你\(N\)个点,\(M\)条带权边的无向图,选出\(K\)条边,使得 ...
随机推荐
- 分享一些Comet开发经验
前言 本comet技术主要用于数据库持久层的 穿越防火墙 远程访问.只要有一台中继网站,任意地点的数据库都能被访问. Comet概念介绍 WebIM.网页的客服.meebo等大家听说过了.最近还有个兄 ...
- web开发中禁止因为网速慢导致重复提交数据
var checkSubmitFlg = false; function check() { if (!checkSubmitFlg) { ...
- 【转】类中如何引用server.MapPath()
转至:http://blog.csdn.net/tangjianft/article/details/5357151 今天在写一个上传图片方法时遇到了两个问题:1.public string getI ...
- c# 范型Dictionary实用例子
//定义 public static Dictionary<string, object> dict =new Dictionary<string, object>(); // ...
- Hadoop Java开发实用快捷键收藏
不断总结更新.... Alt + / 补全 Ctrl + T 打出结构 Ctrl + 2 ,再选择 Quick Assist - Assign to local variable Ctrl ...
- rdlc 分页操作
工具箱中拖一个列表过来,设置 列表-->行组-->组属性常规-->组表达式=Int((RowNumber(Nothing)-1)/10)分页符-->勾选在组的结尾
- C++11右值引用
[C++11右值引用] 1.什么是左值?什么是右值? 左值是表达式结束后依然存在的对象:右值是表达式结束时就不再存在的对象. 2.std::move的作用是什么? std::move用于把任意类型转化 ...
- C#下载http文件
@(编程) using System; using System.IO; using System.Net; namespace Wisdombud.Util { public class HttpH ...
- BootCamp支持软件4/5
按 Mac 机型列出的 Boot Camp 要求 不同的 Mac 电脑适用不同版本的 Windows.如果您不知道您拥有的 Mac 是什么机型,请从 Apple 菜单中选取“关于本机”. 每个表格条目 ...
- Cisco ASA5500系列防火墙恢复IOS全过程
擦除防火墙配置的命令是write erase而不是erase flash!当ASA5510的flash被erase后,如何将新的IOS拷贝到5510内呢? 如下:1. 当flash被erase后设备会 ...