Poj 1279 Art Gallery

Art Gallery

This problem will be judged on PKU. Original ID: 1279
64-bit integer IO format: %lld      Java class name: Main

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

有坑，
首先這些點是順時針的，那麼我讀入這些點之後就要逆置一下，
然後不能排序
第一個半平面交(O(n^2))
代碼：

//(O(n^2))
#include 
#include 
#include 
#define maxn 10000
using namespace std;
typedef long long int LLI;

const double eps = 1e-18;
int sgn(double x) {
    if(fabs(x) < eps)return 0;
    if(x < 0)return -1;
    else return 1;
}
struct Point {
    double x,y;
    Point() {}
    Point(double _x,double _y) {
        x = _x;
        y = _y;
    }
    Point operator -(const Point &b)const {
        return Point(x - b.x, y - b.y);
    }
    double operator ^(const Point &b)const {
        return x*b.y - y*b.x;
    }
    double operator *(const Point &b)const {
        return x*b.x + y*b.y;
    }
};
//計算多邊形面積
double CalcArea(Point p[],int n) {
    double res = 0;
    for(int i = 0; i < n; i++)
        res += (p[i]^p[(i+1)%n]);
    return fabs(res/2);
}
//通過兩點，確定直線方程
void Get_equation(Point p1,Point p2,double &a,double &b,double &c) {
    a = p2.y - p1.y;
    b = p1.x - p2.x;
    c = p2.x*p1.y - p1.x*p2.y;
}
//求交點
Point Intersection(Point p1,Point p2,double a,double b,double c) {
    double u = fabs(a*p1.x + b*p1.y + c);
    double v = fabs(a*p2.x + b*p2.y + c);
    Point t;
    t.x = (p1.x*v + p2.x*u)/(u+v);
    t.y = (p1.y*v + p2.y*u)/(u+v);
    return t;
}

Point tp[110];

void Cut(double a,double b,double c,Point p[],int &cnt) {
    int tmp = 0;
    for(int i = 1; i <= cnt; i++) {
        //當前點在左側，逆時針的點
        if(a*p[i].x + b*p[i].y + c < eps)tp[++tmp] = p[i];
        else {
            if(a*p[i-1].x + b*p[i-1].y + c < -eps)
                tp[++tmp] = Intersection(p[i-1],p[i],a,b,c);
            if(a*p[i+1].x + b*p[i+1].y + c < -eps)
                tp[++tmp] = Intersection(p[i],p[i+1],a,b,c);
        }
    }
    for(int i = 1; i <= tmp; i++)
        p[i] = tp[i];
    p[0] = p[tmp];
    p[tmp+1] = p[1];
    cnt = tmp;
}
const double INF = 100000.0;

Point p[110];
Point List[maxn];

int main() {
//    freopen("in.txt","r",stdin);
    int t;
    scanf("%d",&t);
    while(t --) {
        int n;
        scanf("%d",&n);
        for(int i = 0; i < n; i ++)
            scanf("%lf%lf",&List[i].x,&List[i].y);
        reverse(List,List + n);
        p[1] = Point(-INF,-INF);
        p[2] = Point(INF,-INF);
        p[3] = Point(INF,INF);
        p[4] = Point(-INF,INF);
        p[0] = p[4];
        p[5] = p[1];
        int cnt = 4;
        for(int i = 1; i <= n; i ++) {
            double a,c,b;
            a = List[i % n].y - List[i - 1].y;
            b = -List[i % n].x + List[i - 1].x;
            c = List[i % n].x * List[i - 1].y - List[i - 1].x * List[i % n].y;
            Cut(a,b,c,p,cnt);
        }
        printf("%.2f\n",CalcArea(p,cnt));
    }
    return 0;
}