hduoj 4706 Herding 2013 ACM/ICPC Asia Regional Online —— Warmup

hduoj 4706 Children's Day 2013 ACM/ICPC Asia Regional Online —— Warmup

Herding

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2005    Accepted Submission(s): 563

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：

题意是在N棵树中选出M棵围成的区域面积最小。

换句话说就是求在一堆笛卡尔坐标中选出三个点组成的面积最小。

AC代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#include<stack>
#include<map>
#include<cmath>
#include<string>
using namespace std;
#define N 1000000
struct Point{
    double x;
    double y;
}point[110];
double area2(double x0, double y0 , double x1, double y1, double x2, double y2)
{
    return fabs(x0*y1+x2*y0+x1*y2-x2*y1-x0*y2-x1*y0);
}
int main(){
    int n;
    int tcase;
    scanf("%d", &tcase);
    while(tcase--){
        double minn = 10000005;
        bool flag = false;
        scanf("%d", &n);
        for(int i = 0; i < n; i++){
            scanf("%lf%lf", &point[i].x, &point[i].y);
        }
        if(n >= 3){
            for(int i = 0; i < n-2; i++){
                for(int j = i+1; j < n-1; j++){
                    for(int k = j+1; k < n; k++){
                        double x1, x2, x3, y1, y2, y3;
                        x1 = point[i].x; y1 = point[i].y;
                        x2 = point[j].x; y2 = point[j].y;
                        x3 = point[k].x; y3 = point[k].y;
                        double num = area2(x1, y1, x2, y2, x3, y3);
                        num /= 2.0;
                        if(num < minn && num != 0){
                            minn = num;
                            flag = true;
                        }
                    }
                }
            }
        }
        if(flag)
            printf("%.2lf
", minn);
        else
            printf("Impossible
");
    }
    return 0;
}