UESTC 802 E - Just a Line

Description

There are N points on a plane, among them N−1 points will form a line, your task is to find the point that is not on the line.

Input

The first line contains a single number N, the number of points. (4≤N≤50000)

Then come N lines each with two numbers (xi,yi), giving the position of the points. The points are given in integers. No two points’ positions are the same. (−109≤xi,yi≤109)

Output

Output the position of the point that is not on the line.

Sample Input

5

0 0

1 1

3 4

2 2

4 4

Sample Output

3 4

题意：找出那个不在直线上的点。先算出斜率，再遍历一次。

（排序那一段没用的gg）

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
typedef struct
{
int x;int y;
}Node;
int Sort(const void * a,const void * b)
{
Node *p1=(Node *)a;
Node *p2=(Node *)b;
return p1->x>p2->x;
}
int main (void)
{
Node point[100000];
int t ;
cin>>t;
for(int i=0;i<t;i++)
{
scanf("%d %d",&point[i].x,&point[i].y);
//  printf("%d %d\n",point[i].x,point[i].y);
}

qsort(point,t,sizeof(Node),Sort);
//  printf("after\n");
//  for(int i=0;i<t;i++)
//  {
//      printf("%d %d\n",point[i].x,point[i].y);
//  }

//  printf("-------------\n");
int ans1,ans2;
int tmp1,tmp2;
for(int k=0;k<=3;k++)
{
for(int j=0;j<k;j++)
{
for(int i=0;i<j;i++)
{
if((point[j].y-point[i].y)*(point[k].x-point[i].x)==(point[k].y-point[i].y)*(point[j].x-point[i].x))
{
ans1=point[k].x-point[i].x;
ans2=point[k].y-point[i].y;
tmp1=point[i].x;
tmp2=point[i].y;
break;
}
}
}
}

for(int i=0;i<t;i++)
{
if((point[i].y-tmp2)*ans1!=ans2*(point[i].x-tmp1))

{
printf("%d %d\n",point[i].x,point[i].y);
break;
}
}

return 0;
}