ConvexScore
题目描述
You are given N points (xi,yi) located on a two-dimensional plane. Consider a subset S of the N points that forms a convex polygon. Here, we say a set of points S forms a convex polygon when there exists a convex polygon with a positive area that has the same set of vertices as S. All the interior angles of the polygon must be strictly less than 180°.
For example, in the figure above, {A,C,E} and {B,D,E} form convex polygons; {A,C,D,E}, {A,B,C,E}, {A,B,C}, {D,E} and {} do not.
For a given set S, let n be the number of the points among the N points that are inside the convex hull of S (including the boundary and vertices). Then, we will define the score of S as 2n−|S|.
Compute the scores of all possible sets S that form convex polygons, and find the sum of all those scores.
However, since the sum can be extremely large, print the sum modulo 998244353.
Constraints
1≤N≤200
0≤xi,yi<104(1≤i≤N)
If i≠j, xi≠xj or yi≠yj.
xi and yi are integers.
![](http://exam.upc.edu.cn/upload/image/20180426/20180426085426_81485.jpg)
For a given set S, let n be the number of the points among the N points that are inside the convex hull of S (including the boundary and vertices). Then, we will define the score of S as 2n−|S|.
Compute the scores of all possible sets S that form convex polygons, and find the sum of all those scores.
However, since the sum can be extremely large, print the sum modulo 998244353.
Constraints
1≤N≤200
0≤xi,yi<104(1≤i≤N)
If i≠j, xi≠xj or yi≠yj.
xi and yi are integers.
输入
The input is given from Standard Input in the following format:
N
x1 y1
x2 y2
:
xN yN
N
x1 y1
x2 y2
:
xN yN
输出
Print the sum of all the scores modulo 998244353.
样例输入
4
0 0
0 1
1 0
1 1
样例输出
5
提示
We have five possible sets as S, four sets that form triangles and one set that forms a square. Each of them has a score of 20=1, so the answer is 5.
题意: 平面上有n(n<=200)个点,对于其中能够构成凸多边形的点集S,计算一个得分score,并输出所有这样能构成凸多边形的点集的得分之和。 score=2^(n-|S|), n: 构成此凸多边形的点数及其内部的点数总和;|S|: 构成此凸多边形的点数。 那么就是要对所有的凸包,求其内部的点数所构成的集合个数。 发现太难求了,总不能枚举所有的凸包啊…… 所以我们就从反面来求,求所有点构成的集合个数-不能构成凸包的集合个数 什么样的点不能构成集合? 单点或共线 所以就枚举所有直线就好了
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <bits/stdc++.h> #define ll long long using namespace std; const int N=250; const int p=998244353; int n; struct orz{ int x,y;}a[N]; ll ans; ll poww(ll x,ll y) { ll ret=1; while(y) { if (y&1) ret=ret*x%p; x=x*x%p; y>>=1; } return ret; } bool check(int a,int b,int c,int d) { if (a*d==b*c) return 1; else return 0; } void solve() { for (int i=1;i<=n;i++) { for (int j=i+1;j<=n;j++) { int cnt=0; for (int k=j+1;k<=n;k++) if (check(a[i].x-a[j].x,a[j].x-a[k].x,a[i].y-a[j].y,a[j].y-a[k].y)) cnt++; ans=(ans-poww(2,cnt)+p)%p; } } } int main() { scanf("%d",&n); for (int i=1;i<=n;i++) scanf("%d%d",&a[i].x,&a[i].y); ans=poww(2,n); ans=(ans-1-n+p)%p; solve(); printf("%lld\n",ans%p); return 0; }