hdu 4353 统计点在三角形内的个数

Finding Mine

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1120    Accepted Submission(s): 298

Problem Description

In bitland, there is an area with M gold mines.

As a businessman, Bob wants to buy just a part of the area, which is a simple polygon, whose vertex can only be chosen from N points given in the input (a simple polygon is a polygon without self-intersection). As a greedy man, he wants to choose the part with a lot of gold mines, but unluckily, he is short with money.

Those M gold mines can also be seen as points, but they may be different from those N points. You may safely assume that there will be no three points lying on the same line for all N+M points.

Bob alreadys knows that the price to buy an area is proportional to its size, so he changes his mind. Now he wants to buy a part like this: If the part's size is A, and contains B gold mines, then A/B will be minimum among all the possible parts he can choose. Now, please tell him that minimum number, if all the parts he can choose has B=0, just output -1.

 

Input

First line of the input is a single integer T(T<=30), indicating there are T test cases.
For each test case, the first line is two integers N(3<=N<=200) and M(1<=M<=500), the number of vertexs and the number of mines. Then N lines follows, the i-th line contains two integers x_i,y_i(-5000<=x_i,y_i<=5000), describing the position of the i-th vertex you can choose. Then M lines follow, the i-th line contains two integers x_i,y_i(-5000<=x_i,y_i<=5000), describing the position of the i-th mine.

 

Output

For each case, you should output “Case #k: ” first, where k indicates the case number between 1 and T. Then output the minimum A/B(rounded after the sixth decimal place) or -1.

 

Sample Input

3

3 1

0 0

0 2

3 0

1 1

4 2

0 0

0 5

5 0

2 2

1 2

2 1

3 1

0 0

0 2

2 0

2 2

 

 

Sample Output

Case #1: 3.000000

Case #2: 5.000000

Case #3: -1

Hint

For the second case, we can choose a polygon ( (0,0),(0,5),(2,2),(5,0) ) with A=10 and B=2, if we choose a
triangle ( (0,0),(0,5),(5,0) ), then A=12.5 and B=2.
For the third case, whatever we choose, we can't have a polygon contain the mines.

 

Author

elfness@UESTC_Oblivion

 

Source

2012 Multi-University Training Contest 6

 

题目大意：给一个n可选取的顶点，m个金矿的坐标，求选取一个多边形它的面积与它所含金矿的个数的比值最少，找不到一个含有金矿的多边形的情况输出-1。

解题思路：可以证明一个最小比值的三角形（a/b）与其它三角形合起来形成的多边形肯定要大于a/b(a/b<min{a1/b1,a2/b2,...an/bn}时，a/b<(a+a1+a2..an)/(b+b1+b2+..+bn))。预处理每条直线上方金矿数量，那么三角形所含金矿数就等于abs(num[i][k]-num[i][j]-num[j][k])(画下图就知道了)。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 using namespace std;

 const int maxn=;
 const double eps=1e-;
 const double inf=1000000000.0;
 int n,m,num[maxn][maxn];
 struct Point
 {
     int x,y;
     Point(int x=,int y=):x(x),y(y) {}
 }a[maxn],b[maxn];
 typedef Point Vector;
 Vector operator -(Vector A,Vector B){return Vector(A.x-B.x,A.y-B.y);}
 int Cross(Vector A,Vector B){ return A.x*B.y-A.y*B.x;}//叉积
 inline double min(double a,double b){return a<b?a:b;}
 bool compx(Point a,Point b){return a.x<b.x;}

 void init()
 {
     for(int i=;i<n;i++)
     {
         for(int j=i+;j<n;j++)
         {
             int temp=;
             for(int k=;k<m;k++)
                 if(a[i].x<=b[k].x && b[k].x<a[j].x && Cross(a[j]-a[i],b[k]-a[i])>)
                     temp++;
             num[i][j]=temp;
         }
     }
 }

 int main()
 {
     int i,j,k,t,icase=;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d%d",&n,&m);
         for(i=;i<n;i++) scanf("%d%d",&a[i].x,&a[i].y);
         sort(a,a+n,compx);
         for(i=;i<m;i++) scanf("%d%d",&b[i].x,&b[i].y);
         init();
         double ans=inf;
         for(i=;i<n;i++)
         {
             for(j=i+;j<n;j++)
             {
                 for(k=j+;k<n;k++)
                 {
                     int temp=abs(num[i][k]-num[i][j]-num[j][k]);
                     if(temp==) continue;
                     ans=min(ans,fabs(Cross(a[j]-a[i],a[k]-a[i])/2.0)/temp);
                 }
             }
         }
         if(fabs(ans-inf)<eps) printf("Case #%d: -1\n",++icase);
         else printf("Case #%d: %.6lf\n",++icase,ans);
     }
     return ;
 }