[BZOJ2618] CQOI2006 Convex polygon | Scan Line __ Computational geometry

The area of the convex polygon and seems to have a lot of way, the Net has O (N^2LOGN) algorithm, the nest will only O (N^3LOGN) scan line violence (seems to be able to do O (n^3) is to solve. The endpoints of all triangles and intersections are sorted together so that all the images are divided into several intervals, each area of which is a number of trapezoidal (triangles can also be used as a trapezoid), and then the total length of the median line of the cumulative area on the line. To find the length of the median line with X=mid to get all the points, and then arrange a sequence, recorded through a point of intersection is into a triangle or a triangle, when 0 to add a length, or very good writing. Not how to tune on, but the answer to-eps can live.

#include <cstdio> #include <iostream> #include <cmath> #include <algorithm> #include <cstdio > #include <cmath> #include <iostream> #include <algorithm> #include <memory.h> #define H1 que [Head] #define H2 que[head+1] #define T1 que[tail] #define T2 QUE[TAIL-1] #define N 510 #define EPS 1e-8 using namespace s
td
	struct point{double x,y;
Point () {} point (double x,double y): x (x), Y (y) {}}p[n],p1,p2,h;
	struct line{point p,v;
Double angle;
}a[n],que[n];
int I,j,head,tail,cnt,n,m,tot;
Double ans; 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* (Double k,point a) {return point (K*A.X,K*A.Y);} double operator* (point A,point b) {return a.x*b.y-a.y*b.x;} BOOL CMP (line A,line b) {if (A.angle!=b.angle) return a.angle<b.angle;else return b.v* (A.P-B.P) >0;} point Cross (l
	INE A,line b) {point t=a.p+a.v; Double k1= (A.P-B.P) *b.v,k2=b.v* (T-B.P), k=k1/(K1+K2);
return K*A.V+A.P;
	BOOL Judge (line X1,line X2,line i) {point t=cross (X1,X2);
Return i.v* (T-I.P) >0;
	void push (line x) {while (Head<tail&&!judge (t1,t2,x)) tail--;
	while (Head<tail&&!judge (h1,h2,x)) head++;
Que[++tail]=x;
	} void HPI () {int i;
	Sort (a+1,a+1+tot,cmp);
	a[0].angle=-10.12345;
	for (i=1,cnt=0;i<=tot;i++) if (a[i].angle!=a[i-1].angle) a[++cnt]=a[i];
	que[1]=a[1];head=tail=1;
	for (i=2;i<=cnt;i++) push (A[i]);
	while (Head<tail&&!judge (T1,T2,H1)) tail--;
	while (Head<tail&&!judge (H1,H2,T1)) head++;
	que[tail+1]=que[head];cnt=0;
for (i=head;i<=tail;i++) P[++cnt]=cross (que[i],que[i+1]);
	Double area () {double s=0;
	if (cnt<3) return s;
	P[CNT+1]=P[1];
	for (int i=1;i<=cnt;i++) s+=p[i]*p[i+1];
return Fabs (s)/2;
	int main () {freopen ("2618.in", "R", stdin);
	scanf ("%d", &n);
	tot=0;
		for (i=1;i<=n;i++) {scanf ("%d", &m);
		scanf ("%lf%lf", &p1.x,&p1.y);
		H=P1; for (j=1;j<m;J + +) {scanf ("%lf%lf", &p2.x,&p2.y);
			A[++TOT].P=P1;A[TOT].V=P2-P1;
			A[tot].angle=atan2 (a[tot].v.y,a[tot].v.x);
		P1=P2;
		} p2=h;
		A[++TOT].P=P1;A[TOT].V=P2-P1;
	A[tot].angle=atan2 (a[tot].v.y,a[tot].v.x);
	} HPI ();
printf ("%.3lf", Area ()); }