題鏈：

http://www.lydsy.com/JudgeOnline/problem.php?id=2618

題解：

計算幾何，半平面交。

給出一些凸包，求面積交。

把所有邊都取出來，直接辦平面交就好。

原來dcmp也不能濫用，之前把所有的大於小於比較都用了dcmp函數，導致錯誤。

（以後除了double型判等，其它時候盡量不用dcmp好了）

代碼：

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MAXN 1550
using namespace std;
const double eps=1e-8;
int dcmp(double x){
	if(fabs(x)<=eps) return 0;
	return x<0?-1:1;
}
struct Point{
	double x,y;
	Point(double _x=0,double _y=0):x(_x),y(_y){}
	void Read(){scanf("%lf%lf",&x,&y);}
}D[MAXN];
typedef Point Vector;
struct Line{
	Point s; Vector v; double ang;
	Line(){}
	Line(Point _s,Vector _v):s(_s),v(_v){ang=atan2(v.y,v.x);}
	friend bool operator < (Line A,Line B){return A.ang<B.ang;}
}L[MAXN];
bool operator == (Point A,Point B){return dcmp(A.x-B.x)==0&&dcmp(A.y-B.y)==0;}
Vector operator + (Vector A,Vector B){return Vector(A.x+B.x,A.y+B.y);}
Vector operator - (Point A,Point B){return Vector(A.x-B.x,A.y-B.y);}
Vector operator * (Vector A,double k){return Vector(A.x*k,A.y*k);}
double operator ^ (Vector A,Vector B){return A.x*B.y-A.y*B.x;}
double operator * (Vector A,Vector B){return A.x*B.x+A.y*B.y;}
bool OnRight(Point P,Line A){return (A.v^(P-A.s))<0;}
Point LLI(Line A,Line B){//Line_Line_Intersection
	Vector u=B.s-A.s;
	double t=(u^B.v)/(A.v^B.v);
	return A.s+A.v*t;
}
int HPI(int n){
	static Point QP[MAXN]; static Line QL[MAXN];
	sort(L+1,L+n+1);
	int l=1,r=1,dnt=0; QL[1]=L[1];
	for(int i=2;i<=n;i++){
		while(l<r&&OnRight(QP[r-1],L[i])) r--;
		while(l<r&&OnRight(QP[l],L[i])) l++;
		QL[++r]=L[i];
		if(QL[r].v*QL[r-1].v>0&&dcmp(QL[r].v^QL[r-1].v)==0){
			r--; if(OnRight(QL[r].s,L[i])) QL[r]=L[i];
		}
		if(l<r) QP[r-1]=LLI(QL[r-1],QL[r]);
	}
	while(l<r&&OnRight(QP[r-1],QL[l])) r--;
	if(r-l<=1) return 0;
	QP[r]=LLI(QL[l],QL[r]);
	for(int i=l;i<=r;i++) D[++dnt]=QP[i];
	dnt=unique(D+1,D+dnt+1)-D-1;
	return dnt-(dnt>1&&D[1]==D[dnt]);
}
double PArea(int n){//Polygon_Area
	double S=0; D[n+1]=D[1];
	for(int i=1;i<=n;i++) S+=D[i]^D[i+1];
	return fabs(S)/2;
}
int main(){
	int n,m,dnt=0,p; scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",&m);p=dnt;
		for(int j=1;j<=m;j++){
			D[++dnt].Read();
			if(j>1) L[dnt-1]=Line(D[dnt-1],D[dnt]-D[dnt-1]);
		}
		L[dnt]=Line(D[dnt],D[p+1]-D[dnt]);
	}
	dnt=HPI(dnt);
	printf("%.3lf\n",PArea(dnt));
	return 0;
}

●BZOJ 2618 [Cqoi2006]凸多邊形