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#include#include#include#includeusing namespace std;#define Sgn(x) (((x)<0)?(-1):(1))#define eps 1e-8#define INF 1e10#define Pi (acos(-1.0))double Deg2Rad(double deg) {return (deg*Pi/180.0);}

计算几何模板四（多边形）#include#include#include#include#include#includeusing namespace std;#define Sgn(x) (((x)<0)?(-1):(1))#define eps 1e-8#define INF 1e10#define Pi (acos(-1.0))double Deg2Rad(

TrianglesThis is a simple problem. Given two triangles A and B, you should determine they are intersect, contain or disjoint. (Public edge or point are treated as intersect.)Input First line contains

Hard challengeTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others) Total Submission(s): 345 Accepted Submission(s): 120Problem Description There are n points on th

Dancing Stars on MeTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others) Total Submission(s): 1271 Accepted Submission(s): 702Problem Description The sky was brushe

MindisTime Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 2678 Accepted Submission(s): 524 Special JudgeProblem Description The center coordin

Quoit DesignTime Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 53084 Accepted Submission(s): 13992Problem Description Have you ever played qu

算法实验：//最接近点对#include<iostream>#include<cstdlib>#include<fstream>#include<algorithm>#include<iomanip>#include<map>#include<cmath>#include<ctime>using namespace std ;const int MAXN=1e6+10;cons

计算几何模板三（三角形）#include#include#include#includeusing namespace std;#define Sgn(x) (((x)<0)?(-1):(1))#define eps 1e-8#define INF 1e10#define Pi (acos(-1.0))double Deg2Rad(double deg) {retur

计算几何模板二（矩形、圆）#include#include#include#includeusing namespace std;#define Sgn(x) (((x)<0)?(-1):(1))#define eps 1e-8#define INF 1e10#define Pi (acos(-1.0))double Deg2Rad(double deg) {retu

二维几何基础#include&amp;amp;amp;amp;amp;lt;bits/stdc++.h&amp;amp;amp;amp;amp;gt;using namespace std;const int MAXN = 1e5+5;const int INF = 0x3f3f3f3f;const double eps = 1e-8;const double PI = acos(-1.0);/**atan2(y,x)返回反正切值，[-PI,PI]弧度，求极...