Herding
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 937 Accepted Submission(s): 254
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
Recommend
liuyiding
几何题，可以采用海伦公式，或者行列式
其中海伦公式为：sqrt(l-a)*(l-b)*(l-c)----》个人觉得此处做起来麻烦。。。
所以果断采用行列式....但需要注意精度问题....不然会错很多次的...lz就因为此wa20余次.....说出来都是泪..
代码：
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #define maxn 1e10 6 using namespace std; 7 double area(double *a,double *b,double *c) //运用行列式求面积 8 { 9 double temp=(a[0]*b[1]+a[1]*c[0]+b[0]*c[1])-(c[0]*b[1]+b[0]*a[1]+a[0]*c[1]); 10 return temp<0? -temp:temp; 11 } 12 int main() 13 { 14 double point[105][2],ans; 15 int t,n,i,j,k; 16 scanf("%d",&t); 17 while(t--) 18 { 19 scanf("%d",&n); 20 for(i=0;i<n;i++) 21 scanf("%lf%lf",&point[i][0],&point[i][1]); 22 ans=maxn; 23 for(i=0 ; i<n-2; i++ ) 24 { 25 for(j=i+1;j<n-1;j++) 26 { 27 for(k=j+1;k<n;k++) 28 { 29 double temp=area(point[i],point[j],point[k])/2.0; 30 if(ans>temp&&temp>1e-8) 31 ans=temp; 32 } 33 } 34 } 35 if(n<3||ans<1e-4||ans==maxn) 36 printf("Impossible "); 37 else 38 printf("%.2lf ",ans); 39 } 40 return 0; 41 }