2016-2017 CT S03E05: Codeforces Trainings Season 3 Episode 5 (2016 Stanford Local Programming Contest, Extended) J

题意：给出n个点，要求一个矩形框将（n/2）+1个点框住，要面积最小

解法：先根据x轴选出i->j之间的点，中间的点(包括两边)按照y排序，固定一边X=(x_j-x_i),Y就枚举点两端的Y坐标，细节是注意要取（n/2）+1个点

事实上这样取里面一定符合要求

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
struct P
{
    int x,y;
}H[10000],X[10000];
bool cmd1(P a,P b)
{
    if(a.x==b.x)
    {
        return a.y<b.y;
    }
    else
    {
        return a.x<b.x;
    }
}
bool cmd2(P a,P b)
{
    if(a.y==b.y)
    {
        return a.x<b.x;
    }
    else
    {
        return a.y<b.y;
    }
}
int main()
{
    int t,n;
    cin>>t;
    while(t--)
    {
        int ans=(1<<31)-1;
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            cin>>H[i].x>>H[i].y;
        }
        if(n==1)
        {
            cout<<"0"<<endl;
            continue;
        }
        int num=n/2+1;
        sort(H+1,H+n+1,cmd1);
        for(int i=1;i<=n;i++)
        {
            for(int j=i+1;j<=n;j++)
            {
                int cot=0;
                //枚举ij之间的点，按x坐标
                for(int k=1;k<=n;k++)
                {
                    if(H[k].x>=H[i].x&&H[k].x<=H[j].x)
                    {
                      //  cout<<"A"<<endl;
                        X[++cot]=H[k];
                    }
                }
                sort(X+1,X+1+cot,cmd2);
                for(int k=1;k<=cot;k++)
                {
                  //  cout<<X[k].y<<"A"<<endl;
                }
                int x1=H[i].x;
                int x2=H[j].x;
               // cout<<x1<<"B"<<x2<<endl;
                for(int k=1;k<=cot-num+1;k++)
                {
                    int y1=X[k].y;
                    int y2=X[k+num-1].y;
                   // cout<<y1<<"B"<<y2<<endl;
                    ans=min(ans,(x1-x2)*(y1-y2));
                }
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}