求多邊形的核。用半平面交算法。
/**//*************************************************************************
Author: WHU_GCC
Created Time: 2007-8-25 15:43:03
File Name: pku1279.cpp
Description:
************************************************************************/
#include <iostream>
#include <cmath>
using namespace std;
#define out(x) (cout << #x << ": " << x << endl)
const int maxint = 0x7FFFFFFF;
typedef long long int64;
const int64 maxint64 = 0x7FFFFFFFFFFFFFFFLL;
template <class T> void show(T a, int n) 
{for (int i = 0; i < n; ++i) cout << a[i] << ' '; cout << endl; }
template <class T> void show(T a, int r, int l) {for (int i = 0; i < r; ++i) show(a[i], l); cout << endl; }
#define EPS 1e-10
#define MaxN 3001
struct point
{
double x, y;
};
struct cp
{
int n;
point p[MaxN];
};
point intersectL(double a1, double b1, double c1, double a2, double b2, double c2)
{
point ret;
ret.y = (a1 * c2 - c1 * a2) / (b1 * a2 - a1 * b2);
if (fabs(a2) < EPS)
ret.x = -(b1 * ret.y + c1) / a1;
else
ret.x = -(b2 * ret.y + c2) / a2;
return ret;
}
bool isEqual(point inpA, point inpB)
{
return (fabs(inpA.x - inpB.x) < EPS && fabs(inpA.y - inpB.y) < EPS);
}
double Cross(point inpA, point inpB, point inpC)
{
return (inpB.x - inpA.x) * (inpC.y - inpA.y) - (inpC.x - inpA.x) * (inpB.y - inpA.y);
}
void Get_line(point inpA, point inpB, double &a1, double &b1, double &c1)
{
a1 = inpB.y - inpA.y;
b1 = inpA.x - inpB.x;
c1 = inpA.y * (inpB.x - inpA.x) - inpA.x * (inpB.y - inpA.y);
}
cp cut(point inpA, point inpB, cp incp)
{
cp ret;
point cross;
int i, j;
double t1, t2;
double a1, b1, c1, a2, b2, c2;
ret.n = 0;
for (i = 0; i < incp.n; i++)
{
j = i + 1;
t1 = Cross(inpA, inpB, incp.p[i]);
t2 = Cross(inpA, inpB, incp.p[j]);
if (t1 < EPS && t2 < EPS)
{
ret.p[ret.n++] = incp.p[i];
ret.p[ret.n++] = incp.p[j];
}
else if (t1 > EPS && t2 > EPS)
continue;
else
{
Get_line(inpA, inpB, a1, b1, c1);
Get_line(incp.p[i], incp.p[j], a2, b2, c2);
cross = intersectL(a1, b1, c1, a2, b2, c2);
if (t1 < EPS)
{
ret.p[ret.n++] = incp.p[i];
ret.p[ret.n++] = cross;
}
else
{
ret.p[ret.n++] = cross;
ret.p[ret.n++] = incp.p[j];
}
}
}
if (ret.n == 0) return ret;
for (i = 1, j = 1; i < ret.n; i++)
if (!isEqual(ret.p[i - 1], ret.p[i]))
ret.p[j++] = ret.p[i];
ret.n = j;
if (ret.n != 1 && isEqual(ret.p[ret.n - 1], ret.p[0])) ret.n--;
ret.p[ret.n] = ret.p[0];
return ret;
}
int main()
{
int ca;
int n;
cp input, ret;
for (scanf("%d", &ca); ca--;)
{
scanf("%d", &n);
input.n = n;
for (int i = 0; i < n; i++)
scanf("%lf%lf", &input.p[i].x, &input.p[i].y);
input.p[input.n] = input.p[0];
ret = input;
for (int i = 0; i < input.n; i++)
ret = cut(input.p[i], input.p[i + 1], ret);
double area = 0.0;
for (int i = 0; i < ret.n; i++)
area += ret.p[i].x * ret.p[(i + 1) % n].y - ret.p[(i + 1) % n].x * ret.p[i].y;
printf("%.2lf\n", abs(area / 2.0));
}
return 0;
}