sdut 2884 Painting Cottages 计算几何

Painting Cottages

Time Limit: 2000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

The new cottage settlement is organized near the capital of Flatland. The construction company that is building the settlement has decided to paint some cottages pink and others — light blue. However, they cannot decide
which cottages must be painted which color. The director of the company claims that the painting is nice if there is at least one pink cottage, at least one light blue cottage, and it is possible to draw a straight line in such a way that pink cottages are
at one side of the line, and light blue cottages are at the other side of the line (and no cottage is on the line itself). The main architect objects that there are several possible nice paintings. Help them to find out how many different nice paintings are
there.
 

输入

The first line of the input file contains n — the number of the cottages (1 ≤ n ≤ 300). The following n lines contain the coordinates of the cottages — each line contains two integer numbers xi and yi (−10^4 ≤ xi, yi ≤ 10^4).
 

输出

Output one integer number — the number of different nice paintings of the cottages.
 

示例输入

4
0 0
1 0
1 1
0 1

示例输出

12

提示

Sample.

来源

2014年山东省第五届ACM大学生程序设计竞赛

示例程序

 

提交 

状态 

讨论

给你一个点集求能分割成几种不同的2个空间

可以转化为求点集能有几条直线

利用3点共线的思想 通过gcd来确定 一条直线最小两个点

ACcode：

#include <cstdio>
#include <map>
#define maxn 1005
using namespace std;
inline int gcd(int a,int b){return b==0?a:gcd(b,a%b);}
struct Point{
int x,y;
Point(){}
inline Point(int _x,int _y){
x=_x;
y=_y;
}
inline double lenx(const Point &b)const{
return x-b.x;
}
inline double leny(const Point &b)const{
return y-b.y;
}
}P[maxn];
int main(){
int n;
while(~scanf("%d",&n)){
for(int i=1;i<=n;++i){
int x,y;
scanf("%d%d",&x,&y);
P[i]=Point(x,y);
}
int ans=0;
for(int i=1;i<=n;++i){
map<pair<int ,int>,int> dp;
for(int j=i+1;j<=n;++j){
int x=P[i].lenx(P[j]);
int y=P[i].leny(P[j]);
int z=gcd(x,y);
if(!dp[make_pair(x/z,y/z)])dp[make_pair(x/z,y/z)]=++ans;
}
}
printf("%d\n",ans<<1);
}
return 0;
}