sdut 2884 Painting Cottages 计算几何

最新推荐文章于 2021-12-06 13:40:24 发布

蒟蒻的ACMer

最新推荐文章于 2021-12-06 13:40:24 发布

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本文探讨了一种新颖的算法挑战，即如何确定不同颜色的小屋布局方案数量，这些方案可以通过一条直线将两种颜色的小屋完全分开。文章详细介绍了输入输出格式、示例，并给出了一段实现这一目标的C++代码。

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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

示例输出

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

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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;
}