题目链接
题解
就一个毒瘤卡常题。。写了那么久
设(f[i][s])表示选了集合(s)中的点，最后一个是(i)，进行转移
要先预处理出两点间的点，然后卡卡常就可以过了
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (register int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
#define lbt(x) (x & -x)
#define res register
using namespace std;
const int maxn = 23,maxm = (1 << 21),INF = 1000000000,P = 100000007;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
int X[maxn],Y[maxn];
inline void add(int& x,int y){
x += y;
if (x >= P) x -= P;
}
int n,bin[maxn],f[maxn][maxm],vis[maxn][maxm],a[maxn][maxn];
inline bool onit(int k,int i,int j){
if (X[k] < min(X[i],X[j]) || X[k] > max(X[i],X[j]) || Y[k] < min(Y[i],Y[j]) || Y[k] > max(Y[i],Y[j]))
return false;
return (X[j] - X[i]) * (Y[k] - Y[i]) - (Y[j] - Y[i]) * (X[k] - X[i]) == 0;
}
inline bool is4(int s){
int cnt = 0;
while (s) cnt++,s -= lbt(s);
return cnt >= 4;
}
inline void init(){
for (res int i = 0; i < n; i++)
for (res int j = 0; j < n; j++)
if (j != i)
for (res int k = 0; k < n; k++)
if (k != i && k != j && onit(k,i,j))
a[i][j] |= bin[k];
}
inline void solve(){
int ans = 0,maxv = (1 << n) - 1;
for (int i = 0; i < n; i++) f[i][bin[i]] = 1;
for (res int s = 0; s <= maxv; s++)
for (res int u = 0; u < n; u++)
if (f[u][s]){
for (int j = 0; j < n; j++) if (!(s & bin[j])){
if ((s & a[u][j]) != a[u][j]) continue;
add(f[j][s | bin[j]],f[u][s]);
}
}
for (res int s = 0; s <= maxv; s++) if (is4(s))
for (int i = 0; i < n; i++) add(ans,f[i][s]);
printf("%d
",ans);
}
int main(){
bin[0] = 1; for (res int i = 1; i < maxn; i++) bin[i] = bin[i - 1] << 1;
n = read();
for (res int i = 0; i < n; i++) X[i] = read(),Y[i] = read();
init();
solve();
return 0;
}