ConvexScore

6603: ConvexScore

时间限制: 1 Sec  内存限制: 128 MB
提交: 61  解决: 27
[提交] [状态] [讨论版] [命题人:admin]

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.

输入

The input is given from Standard Input in the following format:
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.

来源/分类

ABC072&ARC082 

思路：

在平面直角坐标系中有一个n元点集U={A_i(x_i,y_i)|1≤i≤n}。考虑以U的子集S中的点为顶点围成的凸多边形P，若这个凸多边形P内（含边界）的点数为k，则这个子集S的权值为f(S)=2^k-|S|。求所有子集S的权值之和（对998,244,353取余）。

定义一个点集上的凸包运算H:G→S。即：平面上的一个点集G，有凸包S=H(G)。

设凸多边形P内的点（除顶点外）构成的集合为T，则|T|=k-|S|。于是，f(S)=2|T|，即f(S)为T的子集个数。设T’是T的一个子集，则：由于集合S=H(S∪T’)，即S为S∪T’的凸包，故集合S∪T’对f(S)的贡献为1。

于是，对于U的一个子集G，若凸多边形P的顶点集为H(G)，则集合G对ans的贡献为1。实际上，就是求总集合有多少子集能构成凸多边形！

AC代码：

#include <bits/stdc++.h>
using namespace std;
const int mod=998244353;
int x[210],y[210],p[210],n,cnt,ans;
int main()
{
    scanf("%d",&n);
    for (int i=0;i<n;i++)
    {
        scanf("%d %d",&x[i],&y[i]);
    }
    p[0]=1;
    for(int i=0;i<n;i++)
    {
         p[i+1]=(p[i]<<1)%mod;
    }
    ans=p[n]-n-1;//要组成凸包，所有情况减去取一个点的情况和减去都不取的情况
    for(int i=0;i<n;i++) //固定两个点，枚举其他点
    {
        for(int j=0;j<i;j++)
        {
            cnt=0;
            for(int k=0;k<j;k++)
            {
                if((x[i]-x[j])*(y[i]-y[k])==(x[i]-x[k])*(y[i]-y[j]))//斜率相等，共线
                {
                    cnt++;
                }
            }
            ans=(ans-p[cnt]+mod)%mod;
        }
    }
    printf("%d\n",ans);
    return 0;
}