题目背景使用过Android 手机的同学一定对手势解锁屏幕不陌生。Android 的解锁屏幕由3X3 个点组成，手指在屏幕上画一条线，将其中一些点连接起来，即可构成一个解锁图案。如下面三个例子所示:题目描述画线时还需要遵循一些规则:连接的点数不能少于4 个。也就是说只连接两个点或者三个点会提示错误。两个点之间的连线不能弯曲。每个点只能“使用”一次，不可重复。这里的“使用”是指手指划过一个点，该点变绿。两个点之间的连线不能“跨过”另一个点，除非那个点之前已经被“使用”过了。对于最后一条规则，参见下图的解释。左边两幅图违反了该规则; 而右边两幅图(分别为2->4-1-3-6 和6->5-4->1->9-2) 则没有违反规则，因为在“跨过”点时，点已经被“使用”过了。现在工程师希望改进解锁屏幕，增减点的数目，并移动点的位置，不再是一个九宫格形状，但保持上述画线的规则不变。请计算新的解锁屏幕上，一共有多少满足规则的画线方案。
输入输出格式
输入格式：输入文件第一行，为一个整数n,表示点的数目。接下来n 行，每行两个空格分开的整数xix_ixi​ 和yiy_iyi​,表示每个点的坐标。输出格式：输出文件共一行，为题目所求方案数除以100000007 的余数。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<ctime>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
#define duke(i,a,n) for(register int i = a;i <= n;++i)
#define lv(i,a,n) for(register int i = a;i >= n;--i)
#define clean(a) memset(a,0,sizeof(a))
const int INF = 1 << 30;
typedef long long ll;
typedef double db;
template <class T>
void read(T &x)
{char c;bool op = 0;while(c = getchar(), c < '0' || c > '9')if(c == '-') op = 1;x = c - '0';while(c = getchar(), c >= '0' && c <= '9')x = x * 10 + c - '0';if(op) x = -x;
}
template <class T>
void write(T x)
{if(x < 0) putchar('-'), x = -x;if(x >= 10) write(x / 10);putchar('0' + x % 10);
}
const int mod = 1e9 + 7;
const int N = 21;
int dp[1 << N][N];
int x[N],y[N],n,ans;
/*void dfs(int now,int u,int len)//zhuangtai.xianzaidedian,changdu
{if(dp[now][u]) return dp[now][u];for(int i = 1;i <= n;i++){if(now & (1 << (i - 1)) == 0){int ok = 0;duke(j,1,n){if(now & (1 << (i - 1)) == 0){if(check(u,j,i) == false){ok = 1;break;}}}if(ok == 0){dfs(now | (1 << (i - 1)),i,len + 1);dp[now][u] += dp[now | (1 << (i - 1))]}}}
}*/
bool check(int a,int b,int c)
{/*if(a == 1 && c == 3 && b == 2)printf("%d %d %d\n",a,b,c);*/if((y[b] - y[a]) * (x[c] - x[b]) != (y[c] - y[b]) * (x[b] - x[a]))return true;if(x[a] > x[b] && x[c] > x[b])return true;if(x[a] < x[b] && x[c] < x[b])return true;if(y[a] > y[b] && y[c] > y[b])return true;if(y[a] < y[b] && y[c] < y[b])return true;return false;
}
void p(int now)
{duke(i,1,n){if((now & (1 << (i - 1))) != 0)printf("1");elseprintf("0");}
}
int main()
{read(n);duke(i,1,n)read(x[i]),read(y[i]);duke(i,0,n - 1){dp[1 << i][i + 1] = 1;}for(int now = 1;now <= (1 << n) - 1;now++){duke(i,1,n){if((now & (1 << (i - 1))) != 0){duke(j,1,n){if((now & (1 << (j - 1))) != 0 && i != j){int ok = 0;duke(k,1,n){if((now & (1 << (k - 1))) == 0)if(check(i,k,j) == false){ok = 1;//cout<<i<<" "<<j<<" "<<k<<endl;break;}}if(ok == 0){dp[now][i] += dp[now - (1 << (i - 1))][j];
//                            p(now),printf(" ");cout<<i<<" "<<j;printf(" ");p(now - (1 << (i - 1)));
//                            puts("");dp[now][i] %= mod;}}}}}int tot = 0;duke(i,0,n - 1){if((now & (1 << i)) != 0)tot++;}if(tot >= 4){duke(i,1,n){if((now & (1 << (i - 1))) != 0)ans += dp[now][i],ans %= mod;}}}
//    ans += 1;
//    ans /= 2;/*duke(i,0,(1 << n) - 1){duke(j,1,n)if((i & (1 << (j - 1))) != 0)p(i),printf(" "),cout<<dp[i][j]<<endl;}*/printf("%d\n",ans);return 0;
}

// luogu-judger-enable-o2
#include<iostream>
#include<cstdio>
#include<cmath>
#include<ctime>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
#define duke(i,a,n) for(register int i = a;i <= n;++i)
#define lv(i,a,n) for(register int i = a;i >= n;--i)
#define clean(a) memset(a,0,sizeof(a))
const int INF = 1 << 30;
typedef long long ll;
typedef double db;
template <class T>void read(T &x)
{char c;bool op = 0;while(c = getchar(), c < '0' || c > '9')if(c == '-') op = 1;x = c - '0';while(c = getchar(), c >= '0' && c <= '9')x = x * 10 + c - '0';if(op) x = -x;
}
template <class T>
void write(T x)
{if(x < 0) putchar('-'), x = -x;if(x >= 10) write(x / 10);putchar('0' + x % 10);
}
const int mod = 1e8 + 7;
const int N = 21;
int dp[1 << N][N];
int x[N],y[N],n,ans;
bool check(int a,int b,int c)
{/*if(a == 1 && c == 3 && b == 2)printf("%d %d %d\n",a,b,c);*/if((y[b] - y[a]) * (x[c] - x[b]) != (y[c] - y[b]) * (x[b] - x[a]))return true;if(x[a] > x[b] && x[c] > x[b])return true;if(x[a] < x[b] && x[c] < x[b])return true;if(y[a] > y[b] && y[c] > y[b])return true;if(y[a] < y[b] && y[c] < y[b])return true;return false;
}
void p(int now)
{duke(i,1,n){if((now & (1 << (i - 1))) != 0)printf("1");elseprintf("0");}
}
vector <int> mp[N][N];
void init(int x,int y)
{for(int i = 1;i <= n;i++){if(check(x,i,y) == false){mp[x][y].push_back(i);}}
}
int main()
{read(n);duke(i,1,n)read(x[i]),read(y[i]);duke(i,0,n - 1){dp[1 << i][i + 1] = 1;}duke(i,1,n)duke(j,1,n)init(i,j);for(int now = 1;now <= (1 << n) - 1;now++){duke(i,1,n){if((now & (1 << (i - 1))) != 0){duke(j,1,n){if((now & (1 << (j - 1))) != 0 && i != j){int ok = 0;duke(f,0,(int)mp[i][j].size() - 1){int k = mp[i][j][f];if((now & (1 << (k - 1))) == 0){ok = 1;//cout<<i<<" "<<j<<" "<<k<<endl;break;}}if(ok == 0){dp[now][i] += dp[now - (1 << (i - 1))][j];
//                            p(now),printf(" ");cout<<i<<" "<<j;printf(" ");p(now - (1 << (i - 1)));
//                            puts("");dp[now][i] %= mod;}}}}}int tot = 0;duke(i,0,n - 1){if((now & (1 << i)) != 0)tot++;}if(tot >= 4){duke(i,1,n){if((now & (1 << (i - 1))) != 0)ans += dp[now][i],ans %= mod;}}}
//    ans += 1;
//    ans /= 2;/*duke(i,0,(1 << n) - 1){duke(j,1,n)if((i & (1 << (j - 1))) != 0)p(i),printf(" "),cout<<dp[i][j]<<endl;}*/printf("%d\n",ans);return 0;
}