POJ1279-Art Gallery

Art Gallery

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 5193 Accepted: 2220

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

170 04 44 79 713 -18 -64 -4

Sample Output

80.00

//AC代码

/*参考博客：http://blog.csdn.net/accry/article/details/6070621此博客对半平面方面知识讲得很全面，通俗题意：利用半平面交求多边形的核多边形的核:它是平面简单多边形的核是该多边形内部的一个点集该点集中任意一点与多边形边界上一点的连线都处于这个多边形内部。就是一个在一个房子里面放一个摄像头能将所有的地方监视到的放摄像头的地点的集合即为多边形的核经常会遇到让你判定一个多边形是否有核的问题*/#include<iostream>#include<queue>#include<cstdio>#include<algorithm>#include<cstring>#include<iomanip>#include<map>#include<cstdlib>#include<cmath>const int INF=0x7fffffff;const double eps=1e-8;const double PI=acos(-1.0);const double inf=1e5;const int Max=2001;#define zero(x) (((x)>0?(x):-(x))<eps)#define mm(a,b) memset(a,b,sizeof(a))using namespace std;int sign(double x){    return (x>eps)-(x<-eps);}typedef struct Node{    double x;    double y;    Node(const double &_x=0, const double &_y=0) : x(_x), y(_y) {}}point;point list[Max],stack[Max];point qq[Max],pp[Max],pnt[Max];int n;int top;int cnt;int curcnt;double xmult(point p0,point p1,point p2){return(p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}double Distance(point p1,point p2)// 返回两点之间欧氏距离{return( sqrt( (p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y) ) );}bool cmp(point p1,point p2){    double temp;    temp=xmult(list[0],p1,p2);    if(temp>0)        return true;    if(temp==0&&(Distance(list[0],p1)<Distance(list[0],p2)))        return true;    return false;}void convex_hull()//凸包模板{    int i;    for(i=1;i<n;i++)    {        point temp;        if((list[i].y<list[0].y)||(list[i].y==list[0].y&&list[i].x<list[0].x))            swap(list[0],list[i]);    }    sort(list+1,list+n,cmp);    top=1;    stack[0]=list[0];    stack[1]=list[1];    for(i=2;i<n;i++)    {        while(top>=1&&xmult(stack[top-1],stack[top],list[i])<=0)            top--;        top++;        stack[top]=list[i];    }}void Init(int n)//初始化{    int i;    for(i=1;i<=n;i++)    {        pp[i]=pnt[i];    }    pp[n+1]=pnt[1];//n个点有n条边    pp[0]=pnt[n];//用于计算相交的点（第n个点在直线左侧并且第一个点在右侧）    //cnt=n;//多边形顶点数}void Init_pp(){    cnt=4;    pp[1] = Node(-inf, inf);    pp[2] = Node(inf, inf);    pp[3] = Node(inf, -inf);    pp[4] = Node(-inf, -inf);    pp[5] = pp[1];    pp[0]=pp[4];}void GetLine(point u,point v,double &a,double &b,double &c)//两点确定一条直线{    a=v.y-u.y;    b=u.x-v.x;    c=v.x*u.y-u.x*v.y;    //cout<<"a、b、c: "<<a<<" "<<b<<" "<<c<<endl;}point Intersect(point u,point v,double a,double b,double c)//求直线切割交于多边形边上的一点{    double q=fabs(a*u.x+b*u.y+c);    double p=fabs(a*v.x+b*v.y+c);    point res;    res.x=((p*u.x+q*v.x)/(q+p));    res.y=((p*u.y+q*v.y)/(q+p));    //cout<<res.x<<" "<<res.y<<endl;    //cout<<"-----------------"<<endl;    return res;}void CutLine(double a,double b,double c)//直线切割模板{    int i;    curcnt=0;    for(i=1;i<=cnt;i++)    {        //cout<<pp[i].x<<" "<<pp[i].y<<endl;        if(sign(a*pp[i].x+b*pp[i].y+c)>=0)//当前顶点在直线的右侧（或者直线上）        {            //cout<<"i  i     :";            //cout<<pp[i].x<<" "<<pp[i].y<<" "<<a<<" "<<b<<" "<<c<<endl;            qq[++curcnt]=pp[i];        }        else        {            if(sign(a*pp[i-1].x+b*pp[i-1].y+c)>0)//前一个顶点在直线的右侧            {                //cout<<"i  i-1     :";                //cout<<pp[i].x<<" "<<pp[i].y<<" "<<pp[i-1].x<<" "<<pp[i-1].y<<a<<" "<<b<<" "<<c<<endl;                qq[++curcnt]=Intersect(pp[i],pp[i-1],a,b,c);            }            if(sign(a*pp[i+1].x+b*pp[i+1].y+c)>0)//同样的，后一个顶点在直线的右侧            {                //cout<<"i  i+1     :";                //cout<<pp[i].x<<" "<<pp[i].y<<" "<<pp[i+1].x<<" "<<pp[i+1].y<<a<<" "<<b<<" "<<c<<endl;                qq[++curcnt]=Intersect(pp[i],pp[i+1],a,b,c);            }        }    }    //cout<<"cnt:"<<cnt<<"curcnt:"<<curcnt<<endl;    for(i=1;i<=curcnt;i++)//更新切割后的多边形    {        pp[i]=qq[i];    }    pp[curcnt+1]=pp[1];    pp[0]=pp[curcnt];    cnt=curcnt;}int main(){    int m,i,j;    int T;    double a,b,c;    //freopen("D:\\in.txt","r",stdin);    cin>>T;    while(T--)    {        cin>>n;        for(i=1;i<=n;i++)        {            cin>>pnt[i].x>>pnt[i].y;        }        pnt[n+1]=pnt[1];//n个点有n条边        Init(n);        Init_pp();        //顺时针方向为切割后多边形        for(i=1;i<=n;i++)        {            //cout<<"-------------"<<endl;            //cout<<"this::  "<<pnt[i].x<<" "<<pnt[i].y<<" "<<pnt[i+1].x<<" "<<pnt[i+1].y<<endl;            GetLine(pnt[i],pnt[i+1],a,b,c);            CutLine(a,b,c);        }        double A=0;        //cout<<"-----------"<<endl;        for(i=1;i<=cnt;i++)        {            //cout<<pp[i].x<<" "<<pp[i].y<<endl;            A+=xmult(Node(0,0),pp[i],pp[i+1]);        }        //cout<<pp[cnt+1].x<<" "<<pp[cnt+1].y<<endl;        cout<<setprecision(2)<<setiosflags(ios::fixed)<<fabs(A)/2.0<<endl;    }    return 0;}