poj 1279 Art Gallery(多边形的核的面积)

Art Gallery Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 3647 Accepted: 1611 Description The art galleries of the new and very futuristic building of the Center for Balkan Cooperation have the form of polygons (not necessarily convex). When a big exhibition is organized, watching over all of the pictures is a big security concern. Your task is that for a given gallery to write a program which finds the surface of the area of the floor, from which each point on the walls of the gallery is visible. On the figure 1. a map of a gallery is given in some co-ordinate system. The area wanted is shaded on the figure 2. Input The number of tasks T that your program have to solve will be on the first row of the input file. Input data for each task start with an integer N, 5 <= N <= 1500. Each of the next N rows of the input will contain the co-ordinates of a vertex of the polygon ? two integers that fit in 16-bit integer type, separated by a single space. Following the row with the co-ordinates of the last vertex for the task comes the line with the number of vertices for the next test and so on. Output For each test you must write on one line the required surface - a number with exactly two digits after the decimal point (the number should be rounded to the second digit after the decimal point). Sample Input 1 7 0 0 4 4 4 7 9 7 13 -1 8 -6 4 -4 Sample Output 80.00 Source Southeastern Europe 2002

#include<cmath>
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
const int mm=2222;
typedef double DIY;
struct point
{
    DIY x,y;
    point(){}
    point(DIY _x,DIY _y):x(_x),y(_y){}
}g[mm];
point MakeVector(point &P,point &Q)
{
    return point(Q.x-P.x,Q.y-P.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(MakeVector(Q,P),MakeVector(Q,R));
}
struct halfPlane
{
    point s,t;
    double angle;
    halfPlane(){}
    halfPlane(point _s,point _t):s(_s),t(_t){}
    halfPlane(DIY sx,DIY sy,DIY tx,DIY ty):s(sx,sy),t(tx,ty){}
    void GetAngle()
    {
        angle=atan2(t.y-s.y,t.x-s.x);
    }
}hp[mm],q[mm];
point IntersectPoint(halfPlane P,halfPlane Q)
{
    DIY a1=CrossProduct(MakeVector(P.s,Q.t),MakeVector(P.s,Q.s));
    DIY a2=CrossProduct(MakeVector(P.t,Q.s),MakeVector(P.t,Q.t));
    return point((P.s.x*a2+P.t.x*a1)/(a2+a1),(P.s.y*a2+P.t.y*a1)/(a2+a1));
}
bool cmp(halfPlane P,halfPlane Q)
{
    if(fabs(P.angle-Q.angle)<1e-8)
        return MultiCross(P.s,P.t,Q.s)>0;
    return P.angle<Q.angle;
}
bool IsParallel(halfPlane P,halfPlane Q)
{
    return fabs(CrossProduct(MakeVector(P.s,P.t),MakeVector(Q.s,Q.t)))<1e-8;
}
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>1e-8)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]))return;
        while(l<r&&MultiCross(hp[i].s,hp[i].t,IntersectPoint(q[r],q[r-1]))>0)--r;
        while(l<r&&MultiCross(hp[i].s,hp[i].t,IntersectPoint(q[l],q[l+1]))>0)++l;
        q[++r]=hp[i];
    }
    while(l<r&&MultiCross(q[l].s,q[l].t,IntersectPoint(q[r],q[r-1]))>0)--r;
    while(l<r&&MultiCross(q[r].s,q[r].t,IntersectPoint(q[l],q[l+1]))>0)++l;
    q[++r]=q[l];
    for(i=l;i<r;++i)
        g[m++]=IntersectPoint(q[i],q[i+1]);
}
int main()
{
    int i,n,m,t;
    double ans;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0;i<n;++i)
            scanf("%lf%lf",&g[i].x,&g[i].y);
        g
=g[0];
        for(i=0;i<n;++i)
        {
            hp[i]=halfPlane(g[i+1],g[i]);
            hp[i].GetAngle();
        }
        HalfPlaneIntersect(n,m);
        ans=0;
        if(m>2)
        {
            g[m]=g[0];
            for(i=0;i<m;++i)
                ans+=CrossProduct(g[i],g[i+1]);
            if(ans<0)ans=-ans;
        }
        printf("%.2lf\n",ans/2.0);
    }
    return 0;
}