# Points in a Plane

### Problem Statement :

```There are N points on an XY plane. In one turn, you can select a set of collinear points on the plane and remove them. Your goal is to remove all the points in the least number of turns. Given the coordinates of the points, calculate two things:

The minimum number of turns (T) needed to remove all the points.
The number of ways to to remove them in T turns. Two ways are considered different if any point is removed in a different turn.
Input Format

The first line contains the number of test cases T. T test cases follow. Each test case contains N on the first line, followed by N lines giving the coordinates of the points.

Constraints

1 <= T <= 50
1 <= N <= 16
0 <= xi,yi <= 100
No two points will have the same coordinates.

Output Format

Output T lines, one for each test case, containing the least number of turns needed to remove all points and the number of ways to do so. As the answers can be large, output them modulo 1000000007.```

### Solution :

```                            ```Solution in C :

In C++ :

#include<iostream>
#include<set>
#include<map>
#include<string>
#include<stdio.h>
#include<sstream>
#include<algorithm>
#include<queue>
#include<cmath>
#include<string.h>
using namespace std ;
#define MOD 1000000007
#define INF (int)1e9
#define MAXN 20
typedef pair<int,int> P ;

int n,pre[MAXN],fac[MAXN],x[MAXN],y[MAXN] ;
int col[MAXN][MAXN] ;

char bit[1 << MAXN] ;
char good[1 << MAXN] ;
char best[1 << MAXN] ;
char valid[1 << MAXN] ;

char vid,id[1 << MAXN] ;
int memo[1 << MAXN] ;
{
if(bit[mask] <= 2) return 1 ;

for(j = 0;j < n;j++) if(mask & 1 << j)
{
nmask ^= 1 << j ;
break ;
}

int ways = 0,can = best[mask] ;
for(int i = nmask;i > 0;i = ((i - 1) & nmask))
{
int k = i | 1 << j ;
if(valid[k] && best[mask ^ k] == can - 1)
{
ways += solve(mask ^ k) ;
if(ways >= MOD) ways -= MOD ;
}
}
}

void generate()
{
for(int tt = 0;tt < 10;tt++)
{
char in[] = "in .txt" ;
in = tt + '0' ;
FILE * fout = fopen(in,"w") ;

int runs = 50 ;
fprintf(fout,"%d\n",runs) ;
for(int j = 0;j < runs;j++)
{
n = rand() % 18 ;
if(tt == 8) n = 18 ;

char vis ;
memset(vis,0,sizeof vis) ;
for(int i = 0;i < n;i++)
{
if(tt < 3 && j < 15)
{
x[i] = rand() % 10 ;
y[i] = rand() % 10 ;
}
else if(tt < 6 && j < 15)
{
x[i] = rand() % 5 ;
y[i] = rand() % 5 ;
}
else if(tt < 10 && j < 15)
{
x[i] = i ;
y[i] = i + (rand() % 5 - 2) ;
if(y[i] < 0) y[i] = i ;
}
else
{
x[i] = rand() % 100 + 1 ;
y[i] = rand() % 100 + 1 ;
}

if(vis[x[i]][y[i]]) { i-- ; continue ; }
vis[x[i]][y[i]] = 1 ;
}

fprintf(fout,"%d\n",n) ;
for(int i = 0;i < n;i++) fprintf(fout,"%d %d\n",x[i],y[i]) ;
}
fclose(fout) ;
}
}

int main()
{
fac = 1 ;
for(int i = 1;i < MAXN;i++) fac[i] = 1LL * i * fac[i - 1] % MOD ;
for(int i = 1;i < 1 << MAXN;i++) bit[i] = bit[i >> 1] + (i & 1) ;
pre = pre = 1 ;
for(int i = 2;i < MAXN;i++)
{
pre[i] = 1LL * pre[i - 2] * (i - 1) % MOD ;
if(i % 2 == 1) pre[i] += pre[i - 1] ;
pre[i] %= MOD ;
}

// generate() ; return 0 ;

int runs ;
scanf("%d",&runs) ;
while(runs--)
{
scanf("%d",&n) ;
for(int i = 0;i < n;i++) scanf("%d%d",&x[i],&y[i]) ;

memset(col,0,sizeof col) ;
for(int k1 = 0;k1 < n;k1++)
for(int k2 = 0;k2 < n;k2++)
{
for(int j = 0;j < n;j++)
{
int area = x[j] * (y[k1] - y[k2]) + x[k1] * (y[k2] - y[j]) + x[k2] * (y[j] - y[k1]) ;
if(area == 0) col[k1][k2] |= 1 << j ;
}
}

for(int i = 0;i < 1 << n;i++)
{
if(bit[i] <= 2) { valid[i] = true ; continue ; }
for(int j = 0;j < n;j++) if(i & 1 << j)
{
int k1 = -1 ;
for(int k = j + 1;k < n;k++) if(i & 1 << k) { k1 = k ; break ; }
if((col[j][k1] | i) == col[j][k1]) valid[i] = true ;
else valid[i] = false ;
break ;
}
}

best = 0 ;
for(int i = 1;i < 1 << n;i++)
{
if(bit[i] == 1) { best[i] = 1 ; continue ; }
int j;
for(j = 0;j < n;j++) if(i & 1 << j) break ;

int cret = n ;
for(int k = j + 1;k < n;k++)
if(i & 1 << k)
cret = min(cret,1 + best[i & ~col[j][k]]) ;
best[i] = cret ;
}

for(int i = 0;i < 1 << n;i++)
{
good[i] = 1 ;
if(bit[i] <= 2) continue ;
int j;
for(j = 0;j < n;j++) if(i & 1 << j) break ;
if(!good[i ^ 1 << j]) { good[i] = 0 ; continue ; }

for(int k = j + 1;k < n;k++)
if(i & 1 << k)
if(bit[i & col[j][k]] > 2)
good[i] = 0 ;
}

int tot = best[(1 << n) - 1] ;
vid++ ;
int ret = solve((1 << n) - 1) ;
ret = 1LL * ret * fac[tot] % MOD ;
printf("%d %d\n",tot,ret) ;
}

return 0 ;
}

In Java :

import java.util.*;
import java.io.*;

class Solution
{
BufferedWriter out;
StringTokenizer token;

int N;
int[] x,y;
int[] dp,dp3;
boolean[] ok;
int[] member;
int mod = 1000000007;

int BitCount(int x)
{
int ret = 0;
while(x > 0)
{
if( (x&1) != 0 ) ret++;
x >>= 1;
}
return ret;
}

boolean collinear(int set)
{
int ctr = 0;
for(int i = 0; set > 0; i++)
{
if( (set&1) != 0 )
member[ctr++] = i;
set >>= 1;
}
if(ctr <= 2)return true;
int a = x[member]-x[member];
int b = y[member]-y[member];
for(int i = 2; i < ctr; i++)
{
int aa = x[member]-x[member[i]];
int bb = y[member]-y[member[i]];
if(aa*b != a*bb)return false;
}
return true;
}

String binary(int x)
{
String ret = "";
for(int i = 0; i < N; i++)
{
if( ((x>>i)&1) == 0) ret = "0"+ret;
else ret = "1"+ret;
}
return ret;
}

void solve() throws IOException
{
long qq = System.currentTimeMillis();
out = new BufferedWriter(new OutputStreamWriter(System.out));
int T = nextInt();
int twoMax = (1<<16);
dp = new int[twoMax];
x = new int;
y = new int;
ok = new boolean[twoMax];
dp3 = new int[twoMax];
member = new int;
ArrayList<Integer> o;
for(int t = 0; t < T; t++)
{
N = nextInt();
int twoN = (1<<N);
for(int i = 0; i < N; i++)
{
x[i] = nextInt();
y[i] = nextInt();
}
o = new ArrayList<Integer>();
for(int i = twoN-1; i > 0; i--)
{
ok[i] = false;
if(collinear(i))
{
ok[i] = true;
}
}
Arrays.fill(dp,-1);
dp = 0;
dp3 = 1;
int m = 0;
for(int i = 0; i < o.size(); i++)
{
int ii = o.get(i);
for(int j = m; j >= 0; j--)
{
if((ii&j) == 0 && dp[j] != -1)
{
m = Math.max(m,j|ii);
if(dp[j|ii] == -1 || dp[j|ii] > 1+dp[j])
{
dp[j|ii] = 1+dp[j];
dp3[j|ii] = (int)(((long)(dp[j]+1)*dp3[j])%mod);
}
else if(dp[j|ii] == 1+dp[j])
{
dp3[j|ii] += ((long)(dp[j]+1)*dp3[j])%mod;
dp3[j|ii] %= mod;
}
}
}
}
out.write(""+ dp[(twoN)-1] + " " + dp3[(twoN)-1]);
out.newLine();
}
out.flush();
out.close();
input.close();

}

int nextInt() throws IOException
{
if(token == null || !token.hasMoreTokens())
return Integer.parseInt(token.nextToken());
}

Long nextLong() throws IOException
{
if(token == null || !token.hasMoreTokens())
return Long.parseLong(token.nextToken());
}

String next() throws IOException
{
if(token == null || !token.hasMoreTokens())
}

public static void main(String[] args) throws Exception
{
new Solution().solve();
}
}

In C :

#include <stdio.h>

#define P 1000000007

long long g=1,p,t,tt,v,kon,a,b;
long long i,j,k,l,m,n,maz,mmaz;

void uloz(long long mam, long long ind, long long vv)
{
if(ind ==n) {mmaz[mam]=1;return;}

uloz(mam, ind+1,vv);

if(vv&(1<<ind)) uloz(mam+(1<<ind),ind+1,vv);

return;
}

void priamka(long long xx, long long yy)
{
long long vv=0,ii;

for(ii=0;ii<n;ii++)
{
vv*=2;

if((p[ii]-p[xx])*(p[yy]-p[xx]) == (p[ii]-p[xx])*(p[yy]-p[xx]))
vv++;
}

if(maz[vv]==0) uloz(0,0,vv);

maz[vv]=1;

return;
}

void pocitaj(long long ind)
{
long long ii,jj,kk,min;

for(ii=0;ii<(1<<n);ii++) {a[ind][ii]=0;a[ind][ii]=0;}

for(ii=0;ii<(1<<n);ii++)
if(a[ind-1][ii] && mmaz[ii])
{
a[ind] = 1;
a[ind] = (a[ind] + a[ind-1][ii])%P;
kon=1;
}

//printf("%lld=kon\n",kon);

if(kon) return;

a[ind]=0;
a[ind]=0;

for(ii=0;ii<(1<<n);ii++)
if(a[ind-1][ii])
{
g++;
while(((1<<min)&ii) == 0) min++;

for(jj=0;jj<l;jj++)
// if(kk=(maz[jj]&ii)) makaj1(ind,ii,kk,0,0);

if(b[kk=(ii&maz[jj])]!=g && (kk&(1<<min)))
{
b[kk]=g;
a[ind][ii^kk] = 1;
a[ind][ii^kk] = (a[ind][ii^kk] + a[ind-1][ii])%P;
}

}

//kon=1;

return;
}

int main()
{

scanf("%lld",&t);
for(tt=0;tt<t;tt++)
{
scanf("%lld",&n);
for(i=0;i<n;i++) scanf("%lld %lld",&p[i],&p[i]);

for(i=0;i<(1<<n);i++) {maz[i]=0;mmaz[i]=0;}

//  for(i=0;i<n;i++) {mmaz[(1<<i)]=1;maz[(1<<i)]=1;}
if(n==1) {mmaz=1;maz=1;}

for(i=0;i<n;i++)
for(j=i+1;j<n;j++)
{
priamka(i,j);
//   printf("%lld %lld -> %lld\n",i,j,priamka(i,j));
}

for(i=0;i<(1<<n);i++) maz[i]=mmaz[i];

l=0;
for(i=1;i<(1<<n);i++)
if(maz[i]) maz[l++] = i;

//printf("%lld\n",l);

//for(i=0;i<l;i++) printf("%lld..\n",maz[i]);

k=0;
for(i=0;i<(1<<n);i++) a[i]=0;
a[(1<<n)-1] = 1;
a[(1<<n)-1] = 1;

kon=0;
i=0;

while(kon==0)
{
i++;
pocitaj(i);

//  for(j=0;j<(1<<n);j++) printf("%lld %lld---> %lld\n",i,j,a[i][j]);

}

v = a[i];
for(j=2;j<=i;j++) v= (v*j)%P;

printf("%lld %lld\n",i,v);
}

return 0;
}```
```

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