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=505;
const double eps=1e-8;
const double inf=1000000000.0;
int n,m,num[maxn][maxn];
struct Point
{
    int x,y;
    Point(int x=0,int y=0):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=0;i<n;i++)
    {
        for(int j=i+1;j<n;j++)
        {
            int temp=0;
            for(int k=0;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])>0)
                    temp++;
            num[i][j]=temp;
        }
    }
}

int main()
{
    int i,j,k,t,icase=0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(i=0;i<n;i++) scanf("%d%d",&a[i].x,&a[i].y);
        sort(a,a+n,compx);
        for(i=0;i<m;i++) scanf("%d%d",&b[i].x,&b[i].y);
        init();
        double ans=inf;
        for(i=0;i<n;i++)
        {
            for(j=i+1;j<n;j++)
            {
                for(k=j+1;k<n;k++)
                {
                    int temp=abs(num[i][k]-num[i][j]-num[j][k]);
                    if(temp==0) 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
",++icase);
        else printf("Case #%d: %.6lf
",++icase,ans);
    }
    return 0;
}