HDU 4709 Herding

Herding

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2186    Accepted Submission(s): 628

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

 

 

题意：有N个点，求最小封闭图形的面积。

分析：很明显三角形面积最小啊！！用海伦公式求面积，然后就是暴力枚举。

不过这里要注意的就是最后保留小数点后面2位，也就是说，答案小于0.01的就视为0。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<string>
#include<iostream>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<map>
#include<stdlib.h>
#include<algorithm>
#define LL __int64
#define FIN freopen("in.txt","r",stdin)
using namespace std;
const int MAXN=200+5;
const double INF=1e9;
const double eps=1e-2;
struct node
{
    double x, y;
} po[MAXN];
int n;

double dis(int a,int b)
{
    return sqrt( (po[a].x-po[b].x) * (po[a].x-po[b].x) + (po[a].y-po[b].y) * (po[a].y-po[b].y) );
}

double calc(int a,int b,int c)
{
    double lenA=dis(a,b);
    double lenB=dis(a,c);
    double lenC=dis(b,c);
    double p=(lenA+lenB+lenC)/2;
    return sqrt(p*(p-lenA)*(p-lenB)*(p-lenC));
}

int main()
{
    int kase;
    scanf("%d",&kase);
    while(kase--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++) scanf("%lf %lf",&po[i].x,&po[i].y);
        double ans=INF;
        bool flag=false;
        for (int i=0;i<n;i++)
        {
            for (int j=i+1;j<n;j++)
            {
                for (int k=j+1;k<n;k++)
                {
                    double tmp=calc(i,j,k);
                    if(fabs(tmp)>=eps)
                    {
                        flag=true;
                        ans=min(ans,tmp);
                    }
                }
            }
        }
        if (!flag) printf("Impossible
");
        else printf("%.2lf
",ans);
    }
}