2019牛客暑期多校训练营（第三场）H.Magic Line

题目链接

题目描述

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

ZYB got N\ N 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.

输入描述:

There are multiple cases. The first line of the input contains a single integer T(1≤T≤10000)T \ (1 \leq T \leq 10000)T (1≤T≤10000) , indicating the number of cases.

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

It is guaranteed that the sum of N\ N N over all cases does not exceed 2×1052 \times 10^52×105 .

输出描述:

For each case, print four integers x1,y1,x2,y2x_1, y_1, x_2, y_2x1,y1,x2,y2 in a line, representing a line passing through (x1,y1)(x_1, y_1)(x1,y1) and (x2,y2)(x_2, y_2)(x2,y2). Obviously the output must satisfy (x1,y1)≠(x2,y2)(x_1,y_1) \ne (x_2,y_2)(x1,y1)=(x2,y2).
The absolute value of each coordinate must not exceed 10910^9109 . It is guaranteed that at least one solution exists. If there are multiple solutions, print any of them.

示例1

输入

输出

-1 999000000 1 -999000001

思维题~
我们对点按 xxx 为第一关键字，yyy 为第二关键字排序，只要把最中间的两个点错位输出即可，错位有个小技巧，我们假设找到的点为 p1p_1p1 和 p2p_2p2，错位很简单，p1.x+1p_1.x+1p1.x+1，然后 p2.x−1p_2.x-1p2.x−1即可，此时我们能保证两个点的连线是能平分所有点的，但此时可能会过图中的点，怎么办呢，我们发现点的绝对值不过 1000，所以只要在加减一个大于1000的值就能避免和点重合的情况，AC代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
pair<int,int>p[1005];
int eps=1000;
main(){int n,t;scanf("%d",&t);while(t--){scanf("%d",&n);for(int i=0;i<n;i++) scanf("%d%d",&p[i].first,&p[i].second);sort(p,p+n);printf("%d %d %d %d\n",p[n/2-1].first+1,p[n/2-1].second-eps,p[n/2].first-1,p[n/2].second+eps);}
}