ZOJ 3720 Magnet Darts 解题报告

BUPT
Summer training 1

题意：随机往磁性矩形平面各个位置投掷磁性质点，质点会被吸到与其距离最近的整数点，相当与质点落在各个整数点的概率是相等的。在平面内部有一多边形，只有当质点落到多边形内部或者边缘才能得分，得分为a*x+b*y。a，b为题目给定，（x，y）是落点坐标。

解法：只要求出落在多边形内部和边缘的点。因为每个点所能吸收的区域是以该点为中心，边长为1的小矩形。落在大矩形顶点的点要*0.25,在边缘的点*0.5。

//Memory: 188 KB
//Time: 680 MS

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
using namespace std;
const double eps = 1e-10;
const int maxn = 35;
int dcmp(double x)
{
if(fabs(x)<eps) return 0;
else return x < 0? -1:1;
}
struct Point{
double x,y;
Point(){};
Point (double a,double b):x(a),y(b){};
Point operator - (const Point &a)const {return Point (x - a.x , y - a.y );}
bool operator == (const Point &a)const
{
return dcmp(x-a.x)==0&&dcmp(y-a.y)==0;
}
void input()
{
scanf("%lf%lf",&x,&y);
}
void output()
{
printf("%.2lf %.2lf\n",x,y);
}
};
typedef Point Vector;
double Cross(Vector a, Vector b)
{
return a.x*b.y-a.y*b.x;
}
double Dot(Vector a, Vector b)
{
return a.x*b.x+a.y*b.y;
}
bool Onseg(Point p, Point a, Point b)
{
return dcmp(Cross(a-p,b-p))==0&&dcmp(Dot(a-p,b-p))<=0;
}
int Ispinpoly(Point p, Point *poly,int n)
{
int wn = 0;
for(int i = 0; i < n; i++)
{
const Point p1 = poly[i],p2 = poly[(i+1)%n];
if(Onseg(p,p1,p2)) return 1;
int k = dcmp(Cross(p2-p1, p-p1));
int d1 = dcmp(p1.y - p.y);
int d2 = dcmp(p2.y - p.y);
if(k > 0 && d1 <= 0 && d2 > 0) wn++;
if(k < 0 && d2 <= 0 && d1 > 0) wn--;
}
if(wn!=0) return 1;
else return 0;
}
double judge(Point p,Point a,Point b)
{
if(p==a||p==b||Point(a.x,b.y)==p||Point(b.x,a.y)==p)
return 4.0;
if(dcmp(p.x-a.x)==0||dcmp(p.x-b.x)==0||dcmp(p.y-a.y)==0||dcmp(p.y-b.y)==0)
return 2.0;
return 1.0;
}
int main()
{
//freopen("in.txt","r",stdin);
Point p,q,poly[maxn],tmp;
int n,a,b,cnt;
while(scanf("%lf%lf",&p.x,&p.y)!=EOF)
{
cnt = 0;
q.input();
scanf("%d%d%d",&n,&a,&b);
for(int i = 0; i < n; i++)
{
poly[i].input();
}
double ans = 0;
for(int i = (int)p.x; i <= (int)q.x; i++)
for(int j = (int)p.y; j <= (int)q.y; j++)
{
tmp.x = i,tmp.y = j;
if(Ispinpoly(tmp,poly,n))
{
//ans+=(a*i+b*j)/judge(tmp,p,q);
double l=max(i-0.5,p.x),r=min(i+0.5,q.x);
double d=max(j-0.5,p.y),u=min(j+0.5,q.y);
ans+=(r-l)*(u-d)*(a*i+b*j);
}
}
printf("%.3f\n",ans/((q.x-p.x)*(q.y-p.y))+eps);

}
return 0;
}