hdu - 4709 - Herding

题意：给出N个点的坐标，从中取些点来组成一个多边形，求这个多边形的最小面积，组不成多边形的输出"Impossible"（测试组数 T <= 25, 1 <= N <= 100,  -1000 <= 坐标Xi, Yi <= 1000）。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4709

——>>面积最小，若有的话，一定是三角形。判断3点是否能组成一个三角形，若用斜率来做，麻烦且可能会有精度误差，用叉积来判断甚好（只需判断两向量的叉积是否为0）。

注意：N可为1、2，这时不能判断三角形。

#include <cstdio>
#include <cmath>
#include <algorithm>
 
using namespace std;
 
const int maxn = 100 + 10;
const double eps = 1e-10;
const double INF = 1 << 30;
 
int N;
 
struct Point{
    double x;
    double y;
    Point(double x = 0, double y = 0):x(x), y(y){}
}p[maxn];
 
typedef Point Vector;
 
Vector operator + (Point A, Point B){
    return Vector(A.x + B.x, A.y + B.y);
}
 
Vector operator - (Point A, Point B){
    return Vector(A.x - B.x, A.y - B.y);
}
 
Vector operator * (Point A, double p){
    return Vector(A.x * p, A.y * p);
}
 
Vector operator / (Point A, double p){
    return Vector(A.x / p, A.y / p);
}
 
double Cross(Vector A, Vector B){
    return A.x * B.y - B.x * A.y;
}
 
double Area2(Point A, Point B, Point C){
    return Cross(B-A, C-A);
}
 
int dcmp(double x){
    if(fabs(x) < eps) return 0;
    else return x < 0 ? -1 : 1;
}
 
void read(){
    scanf("%d", &N);
    for(int i = 0; i < N; i++) scanf("%lf%lf", &p[i].x, &p[i].y);
}
 
void solve(){
    double Min = INF;
    if(N >= 3){
        for(int i = 0; i < N; i++)
        for(int j = i+1; j < N; j++)
            for(int k = j+1; k < N; k++) if(dcmp(Cross(p[j] - p[i], p[k] - p[i]))){
                double temp = fabs(Area2(p[i], p[j], p[k]));
                Min = min(Min, temp);
            }
    }
    if(dcmp(Min - INF) == 0) puts("Impossible");
    else printf("%.2f\n", Min/2);
}
 
int main()
{
    int T;
    scanf("%d", &T);
    while(T--){
        read();
        solve();
    }
    return 0;
}