ZOJ3720 Magnet Darts

哇塞，一道野生的计算几何题诶，题目还看错了两次，一开始以为要求半平面交，后来发现直接暴力枚举就行了。

要注意矩形的四个角都是real number！还有多边形不一定是凸多边形！

计算每个点的期望，首先判断该点是否在多边形内，如果是，需要知道这个点+-0.5的小矩形与大矩形的交的面积，然后根据Ax+By算一下就行了。

#include <cstdio>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <ctime>
#include <climits>
#include <cmath>
#include <iostream>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <list>
#include <queue>
#include <stack>
#include <deque>
#include <algorithm>
using namespace std;

const int maxn = 110;
struct Point
{
	double x, y;
	Point(double x=0, double y=0): x(x), y(y) {}
};
typedef Point Vector;

const double eps = 1e-6;
bool operator < (const Point &a, const Point &b) {return a.x < b.x || (a.x == b.x && a.y <b.y);}
int dcmp(int x) {if (fabs(x)<eps) return 0; else return x < 0 ? -1 : 1;}
bool operator == (const Point &a, const Point &b) {return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;}

Vector operator + (Vector A, Vector B) {return Vector(A.x+B.x, A.y+B.y);}
Vector operator - (Vector A, Vector B) {return Vector(A.x-B.x, A.y-B.y);}

int Dot(Vector A, Vector B) {return A.x*B.x + A.y*B.y;}
int Cross(Vector A, Vector B) {return A.x*B.y - A.y*B.x;}

bool OnSegment(Point P, Point A, Point B)
{
	return dcmp(Cross(A-P, B-P))==0 && dcmp(Dot(A-P, B-P))<=0;
}

int isPointInPolygon(Point P, Point *A, int n)
{
	int w = 0;
	for (int i = 0; i < n; i++)
	{
		if (OnSegment(P, A[i], A[(i+1)%n])) return -1;
		int k = dcmp(Cross(A[(i+1)%n]-A[i], P-A[i]));
		int d1 = dcmp(A[i].y - P.y);
		int d2 = dcmp(A[(i+1)%n].y - P.y);
		if (k>0 && d1<=0 && d2>0) w++;
		if (k<0 && d2<=0 && d1>0) w--;
	}
	if (w != 0) return 1;
	return 0;
}

int n,A,B;
double L,R,U,D;
Point a[maxn],tmp;
double ans;

double f(Point p)
{
    if (isPointInPolygon(p,a,n))
    {
        double r=(min(p.x+0.5,R)-max(p.x-0.5,L))*(min(p.y+0.5,U)-max(p.y-0.5,D));
        return r*(p.x*A+p.y*B);
    }
    return 0;
}

int main()
{
    while (scanf("%lf%lf%lf%lf",&L,&D,&R,&U)==4)
    {
        scanf("%d%d%d",&n,&A,&B);
        for (int i=0;i<n;i++) scanf("%lf%lf",&a[i].x,&a[i].y);
        ans=0;
        for (double x=ceil(L);x<=R;x+=1.0)
            for (double y=ceil(D);y<=U;y+=1.0)
            {
                tmp=Point(x,y);
                ans+=f(tmp);
            }
        printf("%.3lf\n",ans/(R-L)/(U-D));
    }
	return 0;
}