HDU 4709：Herding

Herding

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2418    Accepted Submission(s): 705

Problem Description

Little John is herding his father's cattles. As a lazy boy, he cannot tolerate chasing the cattles all the time to avoid unnecessary omission. Luckily, he notice that there were N trees in the meadow numbered from 1 to N, and calculated their cartesian coordinates
(Xi, Yi). To herding his cattles safely, the easiest way is to connect some of the trees (with different numbers, of course) with fences, and the close region they formed would be herding area. Little John wants the area of this region to be as small as possible,
and it could not be zero, of course.

 

Input

The first line contains the number of test cases T( T<=25 ). Following lines are the scenarios of each test case.

The first line of each test case contains one integer N( 1<=N<=100 ). The following N lines describe the coordinates of the trees. Each of these lines will contain two float numbers Xi and Yi( -1000<=Xi, Yi<=1000 ) representing the coordinates of the corresponding
tree. The coordinates of the trees will not coincide with each other.

 

Output

For each test case, please output one number rounded to 2 digits after the decimal point representing the area of the smallest region. Or output "Impossible"(without quotations), if it do not exists such a region.

 

Sample Input

1
4
-1.00 0.00
0.00 -3.00
2.00 0.00
2.00 2.00

 

Sample Output

2.00

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup

 

迷失在幽谷中的鸟儿，独自飞翔在这偌大的天地间，却不知自己该飞往何方……

题意：给出n个点的坐标，问取出其中任意点围成的区域的最小值！

这个当然是三角形咯！既然给定了点的坐标，直接使用两向量叉乘除以二便可以解决！

#include<stdio.h>
#include<math.h>
#include<iostream>
using namespace std;
#include<string.h>
struct ab
{
    double x,y;
} ab[105];
double solve(int i,int j,int k)
{
    double a=ab[i].x-ab[k].x;
    double b=ab[j].y-ab[k].y;
    double c=ab[i].y-ab[k].y;
    double d=ab[j].x-ab[k].x;
    return fabs(a*b-c*d)/2.0;
}
int main()
{
    int N;
    cin>>N;
    while(N--)
    {
        int n;
        cin>>n;
        for(int i=0; i<n; i++)
            cin>>ab[i].x>>ab[i].y;
        double minn=0xffffffff;
        for(int i=0; i<n; i++)
            for(int j=i+1; j<n; j++)
                for(int k=j+1; k<n; k++)
                {
                    double t=solve(i,j,k);
                    t=fabs(t);
                    if(t<=1e-7)continue;
                    minn=minn>t?t:minn;
                }
        if(minn==0xffffffff)printf("Impossible\n");
        else printf("%.2lf\n",minn);
    }
    return 0;
}