Hdu4709 Triangle Area

HerdingTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission (s): 702 Accepted Submission (s ): 174 Problem DescriptionLittle 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 wocould be herding area. little John wants the area of this region to be as small as possible, and it cocould not be zero, of course. inputThe 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. outputFor 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 Input14-1.00 0.000.00-3.002.00 0.002.00 2.00 Sample Output2.00 Source2013 ACM/ICPC Asia Regional Online -- Warmup analysis: the minimum area is the minimum area of all triangles composed of all points, however, you must note that the three points that make up a triangle cannot be in one line, either horizontal or vertical, or oblique.

 #include<iostream>  #include<cstdio>  #include<cstdlib>  #include<cstring>  #include<string>  #include<queue>  #include<algorithm>  #include<map>  #include<cmath>  #include<iomanip>  #define INF 99999999  using namespace std;    const int MAX=100+10;  double s[MAX][2];    double calculate(int i,int j,int k){      return fabs((s[j][0]-s[i][0])*(s[k][1]-s[i][1])-(s[k][0]-s[i][0])*(s[j][1]-s[i][1]))/2;  }    int main(){      int t,n;      scanf("%d",&t);      while(t--){          scanf("%d",&n);          for(int i=0;i<n;++i)scanf("%lf%lf",&s[i][0],&s[i][1]);          double sum=INF*1.0;          for(int i=0;i<n;++i){              for(int j=i+1;j<n;++j){                  for(int k=j+1;k<n;++k){                      if(s[i][1] == s[j][1] && s[j][1] == s[k][1])continue;                      if((s[j][1]-s[i][1])/(s[j][0]-s[i][0]) == (s[k][1]-s[j][1])/(s[k][0]-s[j][0]))continue;                      sum=min(sum,calculate(i,j,k));                  }              }          }          if(sum == INF*1.0)printf("Impossible\n");          else printf("%.2lf\n",sum);      }      return 0;  }