学习数论 HDU 4709

经过杭师大校赛的打击，明白了数学知识的重要性

开始学习数论，开始找题练手

Herding

 HDU - 4709 

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.

InputThe 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.OutputFor 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

题意：给出几个点的坐标，求能围成的最小的面积。


思路：多边形都能分割成各种三角形，三角形是最小面积，利用叉乘解决，高中知识（枯了，早忘记了），两个向量叉乘的绝对值除以二就能得出三角形面积

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iterator>
#include<sstream>
#include<string>
#include<queue>
#include<vector>
#include<map>
#include<set>

using namespace std;
#define inf 0x3f3f3f3f

int T;
int n;
double ans, temp;
double ac_x, ac_y, bc_x, bc_y;
double x[1010], y[1010];

int main()
{
    cin >> T;
    while (T--)
    {
        ans = inf;
        cin >> n;
        for (int i = 1; i <= n; i++) {
            cin >> x[i];
            cin >> y[i];
        }
        for (int i = 1; i <= n; i++) {
            for(int j=i+1;j<=n;j++){
                for (int k = j + 1; k <= n; k++) {
                    ac_x = x[k] - x[i];
                    ac_y = y[k] - y[i];
                    bc_x = x[k] - x[j];
                    bc_y = y[k] - y[j];
                    temp = fabs(ac_x*bc_y - ac_y * bc_x);
                    if (temp == 0) { continue; }
                    ans = min(ans, temp / 2);
                }
            }
        }
        if (ans == inf) {
            cout << "Impossible" << endl;
        }
        else {
            printf("%.2lf\n", ans);
        }
    }
    return 0;
}