HDU 3365 New Ground (计算几何)
题意:给定点A[0~n-1]和B[0],B[1],A[0]、A[1]映射到B[0]、B[1],求出其余点的映射B[2]~B[n-1]。
析:运用复数类,关键是用模板复数类,一直编译不过,我明明能编译过,交上就不过,只能写一个复数了。。。
代码如下:
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <vector>
#include <map>
using namespace std ;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f3f;
const int maxn = 10000 + 5;
template<class T>
class Complex
{
public:
Complex( ){real=0;imag=0;}
Complex(T r,T i){real=r;imag=i;}
Complex complex_add(Complex &c2);
Complex complex_minus(Complex &c2);
Complex complex_multiply(Complex &c2);
Complex complex_divide(Complex &c2);
T real1();
T imag1(); public:
friend istream &operator >>(istream &is, Complex<T> &p){
cin >> p.real >> p.imag;
return is;
} private:
T real;
T imag;
}; template<class T>
Complex<T> Complex<T>::complex_add(Complex<T> &c2)
{
Complex<T> c;
c.real=real+c2.real;
c.imag=imag+c2.imag;
return c;
} template <class T>
Complex<T> Complex<T>::complex_minus(Complex <T> &c2)
{
Complex <T> c;
c.real=real-c2.real;
c.imag=imag-c2.imag;
return c;
} template <class T>
Complex<T> Complex<T>::complex_multiply(Complex <T> &c2)
{
Complex <T> c;
c.real=real*c2.real-imag*c2.imag;
c.imag=imag*c2.real+real*c2.imag;
return c;
} template <class T>
Complex<T> Complex<T>::complex_divide(Complex <T> &c2)
{
Complex <T> c;
T d=c2.real*c2.real+c2.imag*c2.imag;
c.real=(real*c2.real+imag*c2.imag)/d;
c.imag=(imag*c2.real-real*c2.imag)/d;
return c;
} template <class T>
T Complex<T>::real1(){
return real;
} template <class T>
T Complex<T>::imag1(){
return imag;
} Complex<double> a[maxn], b[2], ans; int main(){
int T, n; cin >> T;
for(int kase = 1; kase <= T; ++kase){
scanf("%d", &n);
double x, y;
for(int i = 0; i < n; ++i){
cin >> a[i]; }
for(int i = 0; i < 2; ++i){
cin >> b[i];
} Complex<double> tmp = (b[1].complex_minus(b[0]));
Complex<double> tmp1 = (a[1].complex_minus(a[0]));
tmp = tmp.complex_divide(tmp1);
printf("Case %d:\n", kase);
for(int i = 0; i < n; ++i){
ans = (a[i].complex_minus(a[0]));
ans = ans.complex_multiply(tmp);
ans = ans.complex_add(b[0]);
printf("%.2lf %.2lf\n", ans.real1(), ans.imag1());
}
}
return 0;
}
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <string>
#include <algorithm>
#include <vector>
#include <map>
#include <complex>
using namespace std ;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f3f;
const int maxn = 10000 + 5;
complex<double> a[maxn], b[2], ans; int main(){
int T, n; cin >> T;
for(int kase = 1; kase <= T; ++kase){
scanf("%d", &n);
for(int i = 0; i < n; ++i)
scanf("%lf %lf", &a[i].real(), &a[i].imag());
scanf("%lf %lf %lf %lf", &b[0].real(), &b[0].imag(), &b[1].real(), &b[1].imag());
complex<double> tmp = (b[1]-b[0])/(a[1]-a[0]);
printf("Case %d:\n", kase);
for(int i = 0; i < n; ++i){
ans = (a[i]-a[0]) * tmp + b[0];
printf("%.2lf %.2lf\n", ans.real(), ans.imag());
}
}
return 0;
}
HDU 3365 New Ground (计算几何)的更多相关文章
- HDU 5130 Signal Interference(计算几何 + 模板)
HDU 5130 Signal Interference(计算几何 + 模板) 题目链接http://acm.hdu.edu.cn/showproblem.php?pid=5130 Descripti ...
- HDU 4063 Aircraft(计算几何)(The 36th ACM/ICPC Asia Regional Fuzhou Site —— Online Contest)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4063 Description You are playing a flying game. In th ...
- hdu 1348:Wall(计算几何,求凸包周长)
Wall Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submis ...
- hdu 1174:爆头(计算几何,三维叉积求点到线的距离)
爆头 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submissi ...
- HDU 5979 Convex【计算几何】 (2016ACM/ICPC亚洲区大连站)
Convex Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Subm ...
- hdu 4631Sad Love Story<计算几何>
链接:http://acm.hdu.edu.cn/showproblem.php?pid=4631 题意:依次给你n个点,每次求出当前点中的最近点对,输出所有最近点对的和: 思路:按照x排序,然后用s ...
- HDU 4606 Occupy Cities (计算几何+最短路+最小路径覆盖)
转载请注明出处,谢谢http://blog.csdn.net/ACM_cxlove?viewmode=contents by---cxlove 题目:给出n个城市需要去占领,有m条线段是障碍物, ...
- 2017 Multi-University Training Contest - Team 9 1003&&HDU 6163 CSGO【计算几何】
CSGO Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Subm ...
- HDU 5839 Special Tetrahedron 计算几何
Special Tetrahedron 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5839 Description Given n points ...
随机推荐
- RESTful处理JSON
@RequestMapping(value = "/dblist", method = RequestMethod.GET) @ResponseBody public Map< ...
- sed命令常见用法
sed -n 'num1p' file 选出行号为num1的行sed -n 'num1,num2p' file 选出num1~num2行sed -n 'num1,$p' file 选出num1行到文件 ...
- shell变量(字符串)间的连接
shell变量(字符串)间的连接 对于变量或者字符串的连接,shell提供了相当简单的做法,比string的连接还要直接. 直接放到一起或用双引号即可. 例如$a, $b,有 c=$a$b c=$a& ...
- json、pickle\shelve模块(超级好用~!)讲解
json.pickle模块讲解 见我前面的文章:http://www.cnblogs.com/itfat/p/7456054.html shelve模块讲解(超级好用~!) json和pickle的模 ...
- JavaScript笔记——事件
事件一般是用于浏览器和用户操作进行交互.最早是 IE 和 Netscape Navigator 中出现, 作为分担服务器端运算负载的一种手段.直到几乎所有的浏览器都支持事件处理.而 DOM2 级规范开 ...
- 记录一下学习Android的小知识
目前要设计即时通讯的整体架构,包括服务端.Android.IOS.PC.平板等等系统,所以需要研究一下手机的实现方式,开始从Android入手,偶尔在这记录下小知识. ADT: 1.页面功能请求结构, ...
- python爬西刺代理
爬IP代码 import requests import re import dauk from bs4 import BeautifulSoup import time def daili(): p ...
- Linux批量“解压”JAR文件
当你需要”解压“很多jar文件时,可以通过很多方式进行,比如下面这种 1,列出每一个jar文件名,逐个展开 for i in $(ls *sour*.jar);do jar xvf $i;done
- SMO是英文SQL Server Management Objects的缩写(上一篇文章的补充)
最近在项目中用到了有关SQL Server管理任务方面的编程实现,有了一些自己的心得体会,想在此跟大家分享一下,在工作中用到了SMO/SQL CLR/SSIS等方面的知识,在国内这方面的文章并不多见, ...
- 查看自己U盘格式
转自:https://zhidao.baidu.com/question/220844418.html 选中U盘后,鼠标右键单击,选项菜单中点击属性,出的的属性窗口中的“常规”项里边就有U盘的基本信息 ...