hdu4709求三角形面积

最新推荐文章于 2025-09-07 20:32:33 发布

转载最新推荐文章于 2025-09-07 20:32:33 发布 · 99 阅读

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本文探讨了一个经典的几何问题：如何通过连接牧场上的树木形成一个尽可能小但不为零的围栏区域，以帮助懒惰的牧童安全地放牧。文章详细解析了问题的算法解决方案，包括如何避免三点共线的情况，以及如何计算并选取最小面积的三角形。

Herding

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 702 Accepted Submission(s): 174

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

分析：求最小面积就是求所有点构成的所有三角形的最小面积，但是要注意选择构成三角形的三个点不能在一条线上，横，竖，斜不能在一条线上

#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;
}