D. Pair Of Lines（数学+思维）

• Problem - D - Codeforces

---

题意：

• 给n个点，判断这个n个点是否能用不多于两条直线全覆盖

---

思路：

• 如果只有不到三个点，那么直接返回”YES“
• 否则，显然任意挑三个点，这三个点有两种情况
• 三个点重合，在一条直线上
  显然这三个点需要用掉一条边，只需要查看剩下来的点能不能只在一条边上
• 三个点组成一个三角形，与上同理，至少用掉两条边来满足这个三角形的覆盖
  所以判断三种可能是否还有剩余的的点未被覆盖，返回结果
• 用计算几何板子需要调整eps精度，开long double（因为点很大）

---

#include<bits/stdc++.h>
#define debug1(a) cout<<#a<<'='<< a << endl;
#define debug2(a,b) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<endl;
#define debug3(a,b,c) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<endl;
#define debug4(a,b,c,d) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<"  "<<#d<<" = "<<d<<endl;
#define debug5(a,b,c,d,e) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<"  "<<#d<<" = "<<d<<"  "<<#e<<" = "<<e<<endl;
#define endl "\n"
#define fi first
#define se second

#define int long long
//#define double long double
//#define int __int128
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;

//#pragma GCC optimize(3,"Ofast","inline")
//#pragma GCC optimize(2)

//常数定义
const double eps = 1e-10;//根据需要调整
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int N = 1e5+10;
int sgn(double x)//符号函数，eps使用最多的地方
{
    if (fabs(x) < eps)
        return 0;
    if (x < 0)
        return -1;
    else
        return 1;
}

//点类及其相关操作
struct Point
{
    double x, y;
    Point() {}
    Point(double _x, double _y) : x(_x), y(_y) {}
    Point operator-(const Point &b) const { return Point(x - b.x, y - b.y); }
    Point operator+(const Point &b) const { return Point(x + b.x, y + b.y); }

    double operator^(const Point &b) const { return x * b.y - y * b.x; } //叉积
    double operator*(const Point &b) const { return x * b.x + y * b.y; } //点积

    bool operator<(const Point &b) const { return x < b.x || (x == b.x && y < b.y); }
    bool operator==(const Point &b) const { return sgn(x - b.x) == 0 && sgn(y - b.y) == 0; }

    Point Rotate(double B, Point P) //绕着点P，逆时针旋转角度B(弧度)
    {
        Point tmp;
        tmp.x = (x - P.x) * cos(B) - (y - P.y) * sin(B) + P.x;
        tmp.y = (x - P.x) * sin(B) + (y - P.y) * cos(B) + P.y;
        return tmp;
    }
};

double dist(Point a, Point b) { return sqrt((a - b) * (a - b)); } //两点间距离
double len(Point a){return sqrt(a.x * a.x + a.y * a.y);}//向量的长度

struct Line
{
    Point s, e;
    Line() {}
    Line(Point _s, Point _e) : s(_s), e(_e) {}

    //两直线相交求交点
    //第一个值为0表示直线重合，为1表示平行,为2是相交
    //只有第一个值为2时，交点才有意义

    pair<int, Point> operator&(const Line &b) const
    {
        Point res = s;
        if (sgn((s - e) ^ (b.s - b.e)) == 0)
        {
            if (sgn((s - b.e) ^ (b.s - b.e)) == 0)
                return make_pair(0, res); //重合
            else
                return make_pair(1, res); //平行
        }
        double t = ((s - b.s) ^ (b.s - b.e)) / ((s - e) ^ (b.s - b.e));
        res.x += (e.x - s.x) * t;
        res.y += (e.y - s.y) * t;
        return make_pair(2, res);
    }
};

//*判断点在直线上的垂点
Point PointToLine(Point P, Line L)
{
    Point result;
    double t = ((P - L.s) * (L.e - L.s)) / ((L.e - L.s) * (L.e - L.s));
    result.x = L.s.x + (L.e.x - L.s.x) * t;
    result.y = L.s.y + (L.e.y - L.s.y) * t;
    return result;
}

bool used[N];
Point points[N];
int n;
bool check(Line a)
{
    //debug1((PointToLine(points[4],a) == points[4]));
    memset(used, 0, sizeof(used));

    for(int i = 0;i < n;i ++)
    {
        if(PointToLine(points[i],a) == points[i])used[i] = 1;
        else used[i] = 0;
    }

    int cnt = 0;
    Point t1,t2;
    
    for(int i = 0;i < n;i ++)if(!used[i])
    {
        //debug2("re",i);
        cnt ++;
        if(cnt == 2)
        {
            t2 = points[i];
            break;
        }
        t1 = points[i];
    }

    
    if(cnt < 2)return 1;
    else{
        Line remainder_line(t1,t2);
        for(int i = 0;i < n;i ++)if(!used[i])
        {
            if(PointToLine(points[i],remainder_line) == points[i]) used[i] = 1;
            else return 0;
        }
    }
    return 1;
}

void solve() 
{
    cin >> n;
    for(int i = 0;i < n;i ++)
    {
        int x,y;cin >> x;cin >> y;
        points[i].x = x; points[i].y = y;
        //debug2(points[i].x,points[i].y);
    }

    if(n <= 3)
    {
        cout << "YES" << endl;
        return ;
    }

    Line l0(points[0],points[1]);
    Line l1(points[1],points[2]);
    Line l2(points[0],points[2]);

    if(check(l0) || check(l1) || check(l2))
    {
        cout << "YES" << endl;
    }else{
        cout << "NO" << endl;
    }

}

signed main()
{
    
    /*
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    */
    int T = 1;//cin >> T;

    while(T--){
        //puts(solve()?"YES":"NO");
        solve();
    }
    return 0;

}
/*

*/