HDU 4709 Herding（叉积求三角形面积）

Herding

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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 2.00

2.00 0.00

 

Sample Output

2.00

 

题目大意：

           john 要在一片森林里放牛，但是 john 怕牛到处乱跑，而他有比较懒，所以他想找几棵树围起来，这要他就可以吃着小菜、喝着小酒在旁边悠哉了，现在他想让你帮他算出围城的圈中最小的面积是多少。

输入：

输入 t 表示有t组测试数据

输入 n ，表示该组数据中有 n 棵树

接下来 n 行分别输入 n 课数的坐标点

        输出：

输出能围成的最小面积，但不能为 0 

思路分析：

要围成的面积最小，那么肯定就是选三棵树了，小于三棵树的围不成，大于三棵的围成的面积肯定比三棵的面积大，所以就算三棵树围就行了，因此，该题到这就有两种方式求，  一种是 ： 一言不合就搜索 (⊙o⊙)…  从 n 棵树种搜三棵树，围成的面积，取最小的。另一种就是：随随便便就暴力  (*@ο@*) ～  ，直接三个for 循环，暴力求解，记录下来最小值。

不过在这道题中还有个坑的地方，精度问题 23333333333333，如果你只先根据点求出每条边的长度，然后再根据海伦公式求面积，那么恭喜你，你将面临的是 WA，我就因为这wa了无数次  ，顿时一脸懵逼O__O "….....    ，最后想想，应该是精度问题，可能中间计算出现精度损失了。 

好了，说了这么多废话，说说这道题用的公式吧，这道题用的是 由三点坐标直接求三角形面积的公式。

设三角形的三个顶点坐标为   A( x1 , y1 )             B( x2 , y 2 )                     C ( x3 , y3 )

                        则 三角形的  面积   S  =   ( x2 - x1 ) * ( y3 - y1)  - ( y2 - y1 ) * ( x3 - x1 )

证明过程我就不写了，大家可以自己回去推导证明一下

附上代码：（搜索）

#include
#include
#include
#define min(x,y)  x > y ? y : x  
double x[5],y[5],xx[300],yy[300],area;
void dfs(int n,int t)
{
	if(t == 4)
	{
		double area1 = (fabs((x[2]-x[1])*(y[3] - y[1]) - (y[2]-y[1])*(x[3]-x[1])))/2;
		if(area1 == 0)
		return;
		area = min(area1,area);
		return ;
	}
	int i;
	for(i = n;i >= 0;i--)
	{
		x[t] = xx[i];
		y[t] = yy[i];
		dfs(i-1,t+1);
	}
}

int main()
{
	int k;
	scanf("%d",&k);
	while(k--)
	{
		memset(x,0,sizeof(x));
		memset(xx,0,sizeof(xx));
		memset(y,0,sizeof(y));
		memset(yy,0,sizeof(yy));
		area = 99999;
		int l;
		scanf("%d",&l);
		int i;
		for(i = 0;i < l;i++)
		{
			scanf("%lf%lf",&xx[i],&yy[i]);
		}
		dfs(l-1,1);
		if(area == 99999)
		printf("Impossible\n");
		else
		printf("%.2lf\n",area);
	}
	return 0;
}