Gym 101047K Training with Phuket's larvae

http://codeforces.com/gym/101047/problem/K

题目：给定n<=2000条绳子，要你找出其中三条，围成三角形，并且要使得围成的三角形面积最小

思路: 考虑一下三角形面积公式1/2a*b*sinO ，那么可以暴力枚举两条边，第三条边要尽量小，为什么呢?因为我要使得O角尽量小。sin值也小了。这是根据小边对小角得到的。所以要找到第一条>a-b的。这样就能使结果最优。这个找可以用二分啦。同理找一个<a+b的

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int maxn = 2e3+20;
double a[maxn];
const double eps = 1e-6;
bool same(double a,double b)
{
    return fabs(a-b)<eps;
}
double S(double a,double b,double c)
{
    double q = (a+b+c)/2;
    return sqrt(q*(q-a)*(q-b)*(q-c));
}
bool check (double a,double b,double c)
{
    if (a-b>=c) return false;
    if (a-c>=b) return false;
    if (b-c>=a) return false;
    return true;
}
void work ()
{
    int n;
    scanf("%d",&n);
    for (int i=1;i<=n;++i)
        scanf("%lf",&a[i]);
    const double gg=-1;
    double ans = gg;
    sort(a+1,a+1+n);
    for (int i=2;i<=n-1;++i)
    {
        for (int j=i+1;j<=n;++j)
        {
            double find = a[j]-a[i];
            int pos = upper_bound(a+1,a+1+n,find)-a;
            if (pos!=n+1&&pos!=j&&pos!=i&&check(a[i],a[j],a[pos]))
            {
                if (ans==-1) ans=S(a[i],a[j],a[pos]);
                else ans=min(ans,S(a[i],a[j],a[pos]));
            }
            find=a[j]+a[i];
            pos = lower_bound(a+1,a+1+n,find)-a;
            if (pos!=n+1&&pos-1!=j&&pos-1!=i&&check(a[i],a[j],a[pos-1]))
            {
                if (ans==-1) ans=S(a[i],a[j],a[pos-1]);
                else ans=min(ans,S(a[i],a[j],a[pos-1]));
            }
        }
    }
    if (same(ans,gg))
    {
        printf ("-1\n");
        return ;
    }
    printf ("%0.10f\n",ans);
    return ;
}

int main()
{
#ifdef local
    freopen("data.txt","r",stdin);
#endif
    int t;
    scanf("%d",&t);
    while(t--) work();
    return 0;
}