poj 2451 Uyuw's Concert（半平面交求面积）
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Uyuw's Concert
Description
Prince Remmarguts solved the CHESS puzzle successfully. As an award, Uyuw planned to hold a concert in a huge piazza named after its great designer Ihsnayish.
The piazza in UDF - United Delta of Freedom’s downtown was a square of [0, 10000] * [0, 10000]. Some basket chairs had been standing there for years, but in a terrible mess. Look at the following graph. 
In this case we have three chairs, and the audiences face the direction as what arrows have pointed out. The chairs were old-aged and too heavy to be moved. Princess Remmarguts told the piazza's current owner Mr. UW, to build a large stage inside it. The stage must be as large as possible, but he should also make sure the audience in every position of every chair would be able to see the stage without turning aside (that means the stage is in the forward direction of their own). To make it simple, the stage could be set highly enough to make sure even thousands of chairs were in front of you, as long as you were facing the stage, you would be able to see the singer / pianist – Uyuw. Being a mad idolater, can you tell them the maximal size of the stage? Input
In the first line, there's a single non-negative integer N (N <= 20000), denoting the number of basket chairs. Each of the following lines contains four floating numbers x1, y1, x2, y2, which means there’s a basket chair on the line segment of (x1, y1) – (x2, y2), and facing to its LEFT (That a point (x, y) is at the LEFT side of this segment means that (x – x1) * (y – y2) – (x – x2) * (y – y1) >= 0).
Output
Output a single floating number, rounded to 1 digit after the decimal point. This is the maximal area of the stage.
Sample Input 3 10000 10000 0 5000 10000 5000 5000 10000 0 5000 5000 0 Sample Output 54166666.7 Hint
Sample input is the same as the graph above, while the correct solution for it is as below:
I suggest that you use Extended in pascal and long double in C / C++ to avoid precision error. But the standard program only uses double. Source
POJ Monthly,Zeyuan Zhu
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题目：http://poj.org/problem?id=2451
题意：话说没怎么看，不过大概是说给个0,0,到10000,10000的正方形，和一些半平面，求划分后剩下的面积的大小
分析：这题数据较大，不过比较简单，只要你的模板是O(nlogn)的复杂度的话，直接套用即可
注意，有关求面积的都要判断下剩下的半平面是否足够3个，要不会wa的，还有套模板的，记得看看数组大小T_T
代码：
#include<cmath>
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int mm=22222;
const double eps=1e-8;
typedef double diy;
struct point
{
diy x,y;
point(){}
point(diy _x,diy _y):x(_x),y(_y){}
}g[mm];
point Vector(point s,point t)
{
return point(t.x-s.x,t.y-s.y);
}
diy CrossProduct(point P,point Q)
{
return P.x*Q.y-P.y*Q.x;
}
diy MultiCross(point P,point Q,point R)
{
return CrossProduct(Vector(Q,P),Vector(Q,R));
}
struct halfPlane
{
point s,t;
diy angle;
halfPlane(){}
halfPlane(point _s,point _t){s=_s,t=_t,angle=atan2(t.y-s.y,t.x-s.x);}
}hp[mm],q[mm];
point Intersection(halfPlane P,halfPlane Q)
{
diy a1=CrossProduct(Vector(P.s,Q.t),Vector(P.s,Q.s));
diy a2=CrossProduct(Vector(P.t,Q.s),Vector(P.t,Q.t));
return point((P.s.x*a2+P.t.x*a1)/(a1+a2),(P.s.y*a2+P.t.y*a1)/(a1+a2));
}
bool IsParallel(halfPlane P,halfPlane Q)
{
return fabs(CrossProduct(Vector(P.s,P.t),Vector(Q.s,Q.t)))<eps;
}
bool cmp(halfPlane P,halfPlane Q)
{
if(fabs(P.angle-Q.angle)<eps)
return MultiCross(P.s,P.t,Q.s)>0;
return P.angle<Q.angle;
}
void HalfPlaneIntersect(int n,int &m)
{
sort(hp,hp+n,cmp);
int i,l=0,r=1;
for(m=i=1;i<n;++i)
if(hp[i].angle-hp[i-1].angle>eps)hp[m++]=hp[i];
n=m,m=0;
q[0]=hp[0],q[1]=hp[1];
for(i=2;i<n;++i)
{
if(IsParallel(q[r],q[r-1])||IsParallel(q[l],q[l+1]))break;
while(l<r&&MultiCross(hp[i].s,hp[i].t,Intersection(q[r],q[r-1]))>0)--r;
while(l<r&&MultiCross(hp[i].s,hp[i].t,Intersection(q[l],q[l+1]))>0)++l;
q[++r]=hp[i];
}
while(l<r&&MultiCross(q[l].s,q[l].t,Intersection(q[r],q[r-1]))>0)--r;
while(l<r&&MultiCross(q[r].s,q[r].t,Intersection(q[l],q[l+1]))>0)++l;
q[++r]=q[l];
for(i=l;i<r;++i)
g[m++]=Intersection(q[i],q[i+1]);
}
int main()
{
int i,n,m;
diy sx,sy,tx,ty,ans;
while(~scanf("%d",&n))
{
for(i=0;i<n;++i)
{
scanf("%lf%lf%lf%lf",&sx,&sy,&tx,&ty);
hp[i]=halfPlane(point(sx,sy),point(tx,ty));
}
hp[n++]=halfPlane(point(0,0),point(1e4,0));
hp[n++]=halfPlane(point(1e4,0),point(1e4,1e4));
hp[n++]=halfPlane(point(1e4,1e4),point(0,1e4));
hp[n++]=halfPlane(point(0,1e4),point(0,0));
HalfPlaneIntersect(n,m);
g[m]=g[0];
ans=0;
if(m>2)
{
for(i=0;i<m;++i)
ans+=CrossProduct(g[i],g[i+1]);
if(ans<0)ans=-ans;
}
printf("%.1lf\n",ans/2);
}
return 0;
}