hdu 4709 Herding

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

-1.00 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

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #define inf 0x3f3f3f3f
 6 #define exd 1e-8
 7 using namespace std;
 8 struct node
 9 {
10     double x,y;
11 };
12 node dd[111];
13 int n;
14 double solve(node a,node b,node c)
15 {
16     return fabs((a.x-c.x)*(b.y-c.y)-(b.x-c.x)*(a.y-c.y))/2.0;
17 }
18 
19 int main()
20 {
21     int t;
22     scanf("%d",&t);
23     while(t--)
24     {
25         scanf("%d",&n);
26         for(int i=0;i<n;i++)
27         {
28             scanf("%lf%lf",&dd[i].x,&dd[i].y);
29         }
30         if(n<=2)
31         {
32             printf("Impossible
");
33             continue;
34         }
35         double MinRes=inf;
36         for(int i=0;i<n;i++)
37         {
38             for(int j=i+1;j<n;j++)
39             {
40                 for(int k=j+1;k<n;k++)
41                 {
42                     double ans=solve(dd[i],dd[j],dd[k]);
43                     if(ans<MinRes && ans>exd)
44                     {
45                         MinRes=ans;
46                     }
47                 }
48             }
49         }
50         if(fabs(MinRes-inf)<exd)
51         {
52             printf("Impossible
");
53         }
54         else
55         {
56             printf("%.2lf
",MinRes);
57         }
58     }
59     return 0;
60 }

View Code

比赛的时候错了好多次，以为题目意思理解错了，比完了，一查解题报告，没想到时自己的三角形求面积那里有错误，不过到现在我还是不知道，为什么用海伦公式（算出三条边的长）会WA，

不过吃一堑长一智，记得下次求N边形面积的时候，果断使用叉积，避免没必要的精度损失，导致没必要的WA，谨记谨记啊！！