#include
#include
#include
#include
#include
#include
#define exp 1e-10
using namespace std;
inline void RD(int &ret)
{
char c;
do
{
c=getchar();
}
while(c<'0'||c>'9');
ret=c-'0';
while((c=getchar())>='0'&&c<='9')
{
ret=ret*10+(c-'0');
}
}
inline void OT(int a)
{
if(a>=10)
{
OT(a/10);
}
putchar(a%10+'0');
}
struct xl
{
double x,y;
}p[22];
int n;
int f2(double x)//判断是否为零
{
if(fabs(x) {
return 0;
}
else
{
if(x<0)
{
return -1;
}
else
{
return 1;
}
}
}
double f1(xl p1,xl p2,xl p0)//判断边缘
{
return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}
double f3(xl p1,xl p2,xl p0)//判断内部
{
return (p1.x-p0.x)*(p2.x-p0.x)+(p2.y-p0.y)*(p1.y-p0.y);
}
bool f(xl z)
{
int i,j=0;
xl p1,p2;
for(i=0;i {
p1=p[i];
p2=p[(i+1)%n];
if(f2(f1(p1,p2,z))==0&&f2(f3(p1,p2,z))<=0)
{
return true;
}
if(f2(f1(p2,z,p1))>0&&f2(p1.y-z.y)<=0&&f2(p2.y-z.y)>0)
{
j++;
}
if(f2(f1(p2,z,p1))<0&&f2(p1.y-z.y)>0&&f2(p2.y-z.y)<=0)
{
j--;
}
}
if(j!=0)
{
return true;
}
else
{
return false;
}
}
int main()
{
xl a,b,z;
int i,j;
double sum,ans,lx,ly,hx,hy,A,B;
while(scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y)!=EOF)
{
RD(n);
scanf("%lf%lf",&A,&B);
for(i=0;i {
scanf("%lf%lf",&p[i].x,&p[i].y);
}
sum=(a.x-b.x)*(a.y-b.y);
ans=0.0;
for(i=a.x;i<=b.x;++i)
{
for(j=a.y;j<=b.y;++j)
{
z.x=i;
z.y=j;
if(f(z)==true)//求区域面积分数
{
lx=max(i-0.5,a.x);
ly=max(j-0.5,a.y);
hx=min(i+0.5,b.x);
hy=min(j+0.5,b.y);
ans+=(hx-lx)*(hy-ly)*(A*i+B*j);
}
}
}
printf("%.3f\n",ans/sum);
}
return 0;
}
F.Final Exam Arrangement

一道水贪心，找到相应区间标记下就行了。