# 【题目】K-colinear Line

UNIQUE VISION Programming Contest 2022（AtCoder Beginner Contest 248）

E - K-colinear Line

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $500$ points

### Problem Statement

You are given $N$ points in the coordinate plane. For each $1\leq i\leq N$, the $i$-th point is at the coordinates $(X_i, Y_i)$.

Find the number of lines in the plane that pass $K$ or more of the $N$ points.
If there are infinitely many such lines, print Infinity.

### Constraints

• $1 \leq K \leq N \leq 300$
• $\lvert X_i \rvert, \lvert Y_i \rvert \leq 10^9$
• $X_i\neq X_j$ or $Y_i\neq Y_j$, if $i\neq j$.
• All values in input are integers.

### Input

Input is given from Standard Input in the following format:

$N$ $K$
$X_1$ $Y_1$
$X_2$ $Y_2$
$\vdots$
$X_N$ $Y_N$

### Output

Print the number of lines in the plane that pass $K$ or more of the $N$ points, or Infinity if there are infinitely many such lines.

5 2
0 0
1 0
0 1
-1 0
0 -1

### Sample Output 1

6

The six lines $x=0$, $y=0$, $y=x\pm 1$, and $y=-x\pm 1$ satisfy the requirement.
For example, $x=0$ passes the first, third, and fifth points.

Thus, $6$ should be printed.

1 1
0 0

### Sample Output 2

Infinity

Infinitely many lines pass the origin.

Thus, Infinity should be printed.

### 源码

#include <bits/stdc++.h>

using namespace std;

int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
const int MAXN = 310;
int N, K;
cin >> N >> K;
int X[MAXN], Y[MAXN];
for (int i = 0; i < N; i++)
cin >> X[i] >> Y[i];
if (K == 1)
{
cout << "Infinity" << endl;
}
else
{
bool vis[MAXN][MAXN]{};
int ans = 0;
for (int i = 0; i < N - 1; i++)
{
for (int j = i + 1; j < N; j++)
{
if (!vis[i][j])
{
vector<int> parallel;
parallel.push_back(i);
parallel.push_back(j);
for (int k = j + 1; k < N; k++)
{
if (((X[j] - X[i]) * (Y[k] - Y[i])) == ((Y[j] - Y[i]) * (X[k] - X[i])))
{
parallel.push_back(k);
}
}
for (int k = 0; k < parallel.size() - 1; k++)
{
for (int l = k + 1; l < parallel.size(); l++)
{
vis[parallel[k]][parallel[l]] = true;
}
}
if (parallel.size() >= K)
ans++;
}
}
}
cout << ans << endl;
}
return 0;
}