H-Magic Line_2019 牛客暑期多校训练营（第三场）

题目连接：

https://ac.nowcoder.com/acm/contest/883/H

Description

There are always some problems that seem simple but is difficult to solve.

ZYB got N distinct points on a two-dimensional plane. He wants to draw a magic line so that the points will be divided into two parts, and the number of points in each part is the same. There is also a restriction: this line can not pass through any of the points.

Help him draw this magic line.

Input

There are multiple cases. The first line of the input contains a single integer \(T(1<=T<=1000)\), indicating the number of cases.

For each case, the first line of the input contains a single even integer \(N (2 <= N <= 1000)\), the number of points. The following \(N\) lines each contains two integers xi,yi |(xi,yi)| <= 1000, denoting the x-coordinate and the y-coordinate of the -th point.

It is guaranteed that the sum of N over all cases does not exceed 2*10^5.

Output

For each case, print four integers \(x_1, y_1, x_2, y_2\) in a line, representing a line passing through \((x_1, y_1)\) and$ (x_2, y_2)$. Obviously the output must satisfy .

The absolute value of each coordinate must not exceed \(10^9\). It is guaranteed that at least one solution exists. If there are multiple solutions, print any of them.

Sample Input

1

4

0 1

-1 0

1 0

0 -1

Sample Output

-1 999000000 1 -999000001

Hint

题意

二维平面上有n个整数坐标的点，求出一条直线将平面上的点分为数量相等的两部分，且线上不能有点，输出线上两个点确定该直线

题解：

先在左下角无穷远处取一质数坐标点(x,y) 对该点和n个点进行极角排序，设排序后中点坐标为(a,b)则这两点连线会将点分为数量相等的两部分，接着取左下角关于中点的对称点(a+a-x, b+b-y),再将该点左移动一格变成(2a-x-1, 2b-y)

则(x,y) (2a-x-1, 2b-y)两点确定的直线就可以分割点为两部分，且线上不会有点

代码

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAX=100005;
const int INF=999999;
typedef long long ll;

int n,top;
struct Node
{
       ll x,y;
}p[MAX],S[MAX];
ll Cross(Node a,Node b,Node c)
{
       return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
}

ll dis(Node a,Node b)
{
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
 bool cmp(Node a,Node b)
{
   ll flag = Cross(p[1],a,b);
   if(flag != 0) return flag > 0;
   return dis(p[1],a) < dis(p[1],b);
}

int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        int n;
        scanf("%d", &n);
        p[1].x = -400000009; p[1].y = -2e3;

        for(int i=1;i<=n;i++)
            scanf("%lld%lld",&p[i+1].x,&p[i+1].y);
        n++;
        sort(p+2, p+1+n, cmp);

        int pos = n/2 + 1;
        ll a = p[pos].x - p[1].x + p[pos].x-1;
        ll b = p[pos].y - p[1].y + p[pos].y;

        printf("%lld %lld %lld %lld\n", p[1].x, p[1].y, a, b);
    }
}