6603: ConvexScore
时间限制: 1 Sec 内存限制: 128 MB
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.
题目大意：
给n个点的坐标， 求n个点的所有子集中能围成的凸多边形的贡献值的和，每个凸多边形的贡献值是 2^（凸多边形内包含的点的个数， 不包括边界）。
分析:
经过观察我们可以发现，每个凸多边形的贡献值，就是固定选构成凸多边形的点， 然后任意选凸多边形内的点，所构成的多边形（不存在凸多边形，这个很容易发现）的个数。 所以最终我们就只求一下这n个点一共能够构成多少个多边形就ok了。 暴力枚举就行了，用总的情况 - 不能构成多边行的情况（直线、单个点、空集）
AC代码：
/*************************************************
Author : NIYOUDUOGAO
Last modified: 2018-07-31 09:07
Email : [email protected]
Filename : t1.cpp
*************************************************/
#include <bits/stdc++.h>
#define mset(a, x) memset(a, x, sizeof(a))
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int mod = 998244353;
const int N = 1e5 + 5;
ll p[205];
struct node {
int x, y;
}a[205];
int main() {
std::ios::sync_with_stdio(false);
cin.tie(NULL), cout.tie(NULL);
p[0] = 1;
for (int i = 1; i <= 200; i++) {
p[i] = (p[i - 1] << 1) % mod;
}
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i].x >> a[i].y;
}
ll res = (p[n] - n - 1 + mod) % mod;
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 ((a[i].x - a[j].x) * (a[i].y - a[k].y) == (a[i].x - a[k].x) * (a[i].y - a[j].y)) {
cnt++;
}
}
res = (res - p[cnt] + mod) % mod;
}
}
cout << res << endl;
return 0;
}