hdu------2488Tornado(几何)
Tornado
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 196 Accepted Submission(s): 48
In summer, tornado often occurs in the place where Professor Jonathan lives. After several years of research, Wind Chaser found many formation rules and moving patterns of tornados. In the satellite image, a tornado is a circle with radius of several meters to several kilometers. And its center moves between two locations in a straight line, back and forth at a fixed speed. After observing a tornado’s movement, Wind Chaser will pick a highway, which is also a straight line, and chase the tornado along the highway at the maximum speed of his car.
The smallest distance between the Wind Chaser and the center of the tornado during the whole wind chasing process, is called “observation distance”. Observation distance is critical for the research activity. If it is too short, Wind Chaser may get killed; and if it is too far, Wind Chaser can’t observe the tornado well. After many times of risk on lives and upset miss, Wind Chaser turns to you, one of his most brilliant students, for help. The only thing he wants to know is the forthcoming wind chasing will be dangerous, successful or just a miss.
xw1 yw1 xw2 yw2 vw
xt1 yt1 xt2 yt2 vt
dl du
In the first line, (xw1, yw1) means the start position of Wind Chaser; (xw2, yw2) is another position in the highway which Wind Chaser will definitely pass through; and vw is the speed of the car. Wind chaser will drive to the end of the world along that infinite long highway.
In the second line, (xt1, yt1) is the start position of tornado; (xt2, yt2) is the turn-around position and vt is the tornado’s speed. In other words, the tornado’s center moves back and forth between (xt1, yt1) and (xt2, yt2) at speed vt .
The third line shows that if the observation distance is smaller than dl , it will be very dangerous; and if the observation distance is larger than du, it will be a miss; otherwise it will lead to a perfect observation.
All numbers in the input are floating numbers.
-2000000000 <= xw1, yw1, xw2, yw2, xt1, yt1, xt2, yt2 <= 2000000000
1 <= vw, vt <= 20000
0 <= dl, du <= 2000000
Note:
1. It’s guaranteed that the observation distance won’t be very close to dl or du during the whole wind chasing process. There will be at least 10-5 of difference.
2. Wind Chaser and the tornado start to move at the same time from their start position.
10 -5 12 7 4
1.3 2.7
0 0 1 0 2
10 -5 12 7 1
0.3 0.4
Perfect
代码:
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
const int MAX= ;
const double esp = 1e-; struct Tnode
{
double x,y;
}w1,w2,t1,t2;
double vw,vt,dl,du;
double getmin(double a , double b)
{
return (a>b)?b:a;
}
//求点积
double dianji(Tnode &a ,Tnode &b ,Tnode &c)
{
return (b.x-a.x)*(c.x-a.x)+(b.y-a.y)*(c.y-a.y);
}
//求叉 积
double det(Tnode &a ,Tnode &b ,Tnode &c)
{
return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);
}
//求距离
double dis(Tnode &a,Tnode &b)
{
return sqrt(fabs((b.x-a.x)*(b.x-a.x)+(b.y-a.y)*(b.y-a.y)));
}
//求点o到线段的最近的距离
double getdistance(Tnode &o,Tnode a ,Tnode b,double dx,double dy)
{
a.x+=dx;
a.y+=dy;
b.x+=dx;
b.y+=dy;
double d =getmin(dis(o,a),dis(o,b));
double di=dis(a,b);
if(di<=esp) return dis(o,a);
if(dianji(a,o,b)>=-esp&&dianji(b,o,a)>=-esp)
return fabs(det(a,b,o))/di;
else
return getmin(dis(o,a),dis(o,b));
}
//求o到以线段ab为起始,(dx,dy)为间距的平行线段的最近距离
double calc(Tnode &o ,Tnode &a ,Tnode &b , double dx, double dy)
{
Tnode a1,b1;
int ll=,rr=MAX;
while(ll<rr)
{
int mid=(ll+rr)/;
double d1=getdistance(o,a,b,dx*mid,dy*mid);
double d2=getdistance(o,a,b,dx*(mid+),dy*(mid+));
if(d1<=d2+esp) rr=mid;
else ll=mid+;
}
return getdistance(o,a,b,dx*ll,dy*ll);
}
void work()
{
Tnode wdr ,tdr,move,a1,b1,a2,b2;
double distance,time,d,d1,d2;
distance=dis(w1,w2);
wdr.x = (w2.x-w1.x)*vw/distance;
wdr.y = (w2.y-w1.y)*vw/distance;
distance = dis(t1,t2);
time = distance/vt;
tdr.x = (t2.x-t1.x)*vt/distance;
tdr.y=(t2.y-t1.y)*vt/distance;
move.x=(-wdr.x+tdr.x)*time;
move.y=(-wdr.y+tdr.y)*time;
//求两个线段簇的第一条线段a1-b1和a2-b2
a1=t1;
b1.x=a1.x+move.x;
b1.y=a1.y+move.y;
move.x=(-wdr.x-tdr.x)*time;
move.y=(-wdr.y-tdr.y)*time;
a2=b1;
b2.x=a2.x+move.x;
b2.y=a2.y+move.y;
//分别求点w1到两个线段簇的最近距离d1和d3
d1=calc(w1,a1,b1,b2.x-a1.x,b2.y-a1.y);
d2=calc(w1,a2,b2,b2.x-a1.x,b2.y-a1.y);
//判断结果
d=getmin(d1,d2);
if(d+esp<d1) printf("Dangerous\n");
else if(d-esp>du) printf("Miss\n");
else printf("Perfect\n");
}
int main()
{
while(scanf("%lf",&w1.x)!=EOF)
{
scanf("%lf%lf%lf%lf",&w1.y,&w2.x,&w2.y,&vw);
scanf("%lf%lf%lf%lf%lf",&t1.x,&t1.y,&t2.x,&t2.y,&vt);
scanf("%lf%lf",&dl,&du);
work();
}
return ;
}
hdu------2488Tornado(几何)的更多相关文章
- hdu 5430(几何)
题意:求光在圆内反射n次后第一次返回原点的方案数 如果k和n-1可约分,则表明是循环多次反射方案才返回原点. #include <iostream> #include <cstrin ...
- hdu 1577 WisKey的眼神 (数学几何)
WisKey的眼神 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total S ...
- POJ 3831 & HDU 3264 Open-air shopping malls(几何)
题目链接: POJ:id=3831" target="_blank">http://poj.org/problem?id=3831 HDU:http://acm.h ...
- HDU 1700 Points on Cycle (几何 向量旋转)
http://acm.hdu.edu.cn/showproblem.php?pid=1700 题目大意: 二维平面,一个圆的圆心在原点上.给定圆上的一点A,求另外两点B,C,B.C在圆上,并且三角形A ...
- HDU 1432 Lining Up(几何)
http://acm.hdu.edu.cn/showproblem.php?pid=1432 题目大意: 2维平面上给定n个点,求一条直线能够穿过点数最多是多少. 解题思路: 因为题目给定的n(1~7 ...
- HDU 1392 Surround the Trees(几何 凸包模板)
http://acm.hdu.edu.cn/showproblem.php?pid=1392 题目大意: 二维平面给定n个点,用一条最短的绳子将所有的点都围在里面,求绳子的长度. 解题思路: 凸包的模 ...
- hdu 5839(三维几何)
Special Tetrahedron Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Othe ...
- hdu 1115 Lifting the Stone (数学几何)
Lifting the Stone Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others ...
- hdu 1086 You can Solve a Geometry Problem too (几何)
You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/3 ...
- HDU 5128.The E-pang Palace-计算几何
The E-pang Palace Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Othe ...
随机推荐
- Cheatsheet: 2014 05.01 ~ 05.31
Web Choosing a Web Framework/Language Combo for the Next Decade Optimizing NGINX and PHP-fpm for hig ...
- java虚拟机能并发的启动多少个线程
新建一个类,导入如下的测试代码: public class TestNativeOutOfMemoryError { public static void main(String[] args) { ...
- 部署PDA程序的时候存储不足的解决办法
通常Windows Mobile的存储空间分为程序内存和存储内存,默认都比较小,当程序比较大的时候可能无法正常部署到设备上面,针对此问题可采用如下办法解决:1.通过VS修改,工具--选项--设备--选 ...
- installing a 3D printer
托公司的福,今天可以自己组装一台3D打印机.心里颇有一种开箱有益的兴奋. 落入手中的是一台Panowin F1,价格不贵,却同时拥有了3D打印功能和激光打印功能.颇有一种小型创客作坊的雏形. 硬件搭建 ...
- hdu 3535 AreYouBusy 分组背包
AreYouBusy Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Probl ...
- 关于Docker在测试方面的应用
Docker 火了很长一段时间了,前段时间简单的学习和试玩了一下子,发现他对测试很有价值,觉得有必要再次深入研究. 这里标记一些较好的学习网址,用作参考: InfoQ上面有系列的文章: 深入浅出Doc ...
- css 集锦。
可以是 链接的下划线 去掉. a {text-decoration: none} position:absolute 绝对定位 position:relative 相对定位 ie 图片失真 -ms ...
- S3C2440的GPIO编程
一.初步认识S3C2440A [S3C2440A简介] S3C2440A是三星公司推出的基于ARM920t内核的32/16位RISC微处理器.主要用于手持设备和中高端电子产品中.它内部集成16k数据c ...
- [转载]bigtable 中文版
转载厦门大学林子雨老师的译文 原文: http://dblab.xmu.edu.cn/post/google-bigtable/ Google Bigtable (中文版) 林子雨2012-05-08 ...
- SQL 总结
1. select 使用正则表达式 正则表达式的模式串, 与linux基本相同, oracle提供以下4个函数来支持正则表达式: REGEXP_LIKE: 比较一个字符串是否与正则表达式匹配(看来是返 ...