Convex hull area of intersection, area and.

Title: Nbut 1640: The public part of the polygon. Test instructions is to ask: whether two simple polygons have intersect. Idea: Calculate the area intersection or area of two convex hull, can solve this problem.
Because the topic is a simple polygon, and the output is counter-clockwise, all do not need to require convex package. To find the code of the area intersection:

#include <stdio.h> #include <string.h> #include <math.h> #include <queue> #include <vector > #include <iostream> #include <complex> #include <string> #include <set> #include <map
> #include <algorithm> using namespace std; #pragma COMMENT (linker, "/uack:1024000000,1024000000") #define LL Long long #define ULL unsiged Long long #define PB push
_back #define MP Make_pair #define MOD 1000003 #define EPS 0.00000001 typedef pair<int, int> PII;
CONST LL INF = 1ll<<50;
const int NN = 100100;

Const double PI = ACOs (-1.0);
	int dcmp (double x) {if (Fabs (x) < EPS) return 0; return x > 0?
1:-1;
	} struct Point {double x, y;
Point () {}-point (double _x, double _y): X (_x), Y (_y) {}};
Point operator + (point A, point B) {return point (a.x + b.x, A.Y + b.y);}
Point operator-(point A, point B) {return point (a.x-b.x, a.y-b.y);}
Point operator * (Point A, double b) {return point (a.x*b,a.y*b);} Point operator/(POint A, double b) {return point (a.x/b, a.y/b);} double Dot (point A, point B) {return a.x*b.x + a.y*b.y;} double Len (Point a) {return sqrt (Dot (A, a)), double Cross (point A, point B) {return a.x*b.y-a.y*b.x;} double Area (vector
	&LT;POINT&GT;P)//Calculate the area of the convex hull {double ans = 0;
	int sz = P.size ();
	for (int i = 1; i < sz-1; i++) ans + = Cross (P[i]-p[0], P[i + 1]-p[0]);
return ans/2.0;
	} Double seg (Point O, point A, point B) {if (dcmp (b.x-a.x) = = 0) return (O.Y-A.Y)/(B.Y-A.Y);
Return (o.x-a.x)/(b.x-a.x);
} vector<point> pp[110];
Pair<double, int> s[2000200];
	Double polyunion (vector<point>*p, int n)//the area intersection of n multiple convex packets {DOUBLE ret = 0;
		for (int i = 0; i < n; i++) {int sz = p[i].size ();
			for (int j = 0; J < Sz; J + +) {int m = 0;
			s[m++] = MP (0, 0);
			s[m++] = MP (1, 0);
			Point A = P[i][j], B = p[i][(j + 1)% SZ];
					for (int k = 0; k < n; k++) {if (i! = k) {int siz = P[k].size (); for (int ii = 0; II < siz;
						ii++) {point c = p[k][ii], d = p[k][(ii + 1)% siz];
						int C1 = DCMP (Cross (b-a, c-a));
						int C2 = dcmp (Cross (b-a, d-a)); if (C1 = = 0 && C2 = = 0) {if (dcmp (Dot (b-a, d-c)) > 0 && i > k) {s[
								m++] = MP (SEG (c, a, b), 1);
							s[m++] = MP (SEG (d, a, b),-1);
							}} else {double S1 = Cross (d-c, a-c);
							Double s2 = Cross (d-c, b-c);
							if (C1 >= 0 && C2 < 0) s[m++] = MP (S1/(S1-S2), 1);
						else if (C1 < 0 && C2 >= 0) s[m++] = MP (S1/(S1-S2),-1);
			}}}} sort (s, S + M);
			Double pre = MIN (max (S[0].first, 0.0), 1.0), now;
			Double sum = 0;
			int cov = S[0].second;
				for (int j = 1; j < M; J + +) {now = min (max (S[j].first, 0.0), 1.0);
				if (!cov) sum + = Now-pre;
				CoV + = S[j].second;
			Pre = now;
		} ret + = Cross (A, b) *sum;
}} return RET/2; } int Main() {int n, m, kcase = 1;
		while (scanf ("%d", &n)! = EOF) {pp[0].clear ();
		Pp[1].clear ();
		Point XP;
			for (int i = 0; i < n; i++) {scanf ("%lf%lf", &xp.x, &AMP;XP.Y);
		PP[0].PB (XP);
		} scanf ("%d", &m);
			for (int i = 0; i < m; i++) {scanf ("%lf%lf", &xp.x, &AMP;XP.Y);
		PP[1].PB (XP);
		} Double t1 = area (pp[0]) + area (pp[1]);
		Double t2 = polyunion (PP, 2);
		printf ("Case%d:", kcase++);
		if (dcmp (t1-t2) = = 0) puts ("No");
	Else puts ("Yes");
} return 0;
 }