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）

  1 #include<bits/stdc++.h>
  2 #define debug1(a) cout<<#a<<'='<< a << endl;
  3 #define debug2(a,b) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<endl;
  4 #define debug3(a,b,c) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<endl;
  5 #define debug4(a,b,c,d) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<"  "<<#d<<" = "<<d<<endl;
  6 #define debug5(a,b,c,d,e) cout<<#a<<" = "<<a<<"  "<<#b<<" = "<<b<<"  "<<#c<<" = "<<c<<"  "<<#d<<" = "<<d<<"  "<<#e<<" = "<<e<<endl;
  7 #define endl "\n"
  8 #define fi first
  9 #define se second
 10 
 11 #define int long long
 12 //#define double long double
 13 //#define int __int128
 14 using namespace std;
 15 
 16 typedef long long LL;
 17 typedef unsigned long long ULL;
 18 typedef pair<int,int> PII;
 19 typedef pair<LL,LL> PLL;
 20 
 21 //#pragma GCC optimize(3,"Ofast","inline")
 22 //#pragma GCC optimize(2)
 23 
 24 //常数定义
 25 const double eps = 1e-10;//根据需要调整
 26 const double PI = acos(-1.0);
 27 const int INF = 0x3f3f3f3f;
 28 const int N = 1e5+10;
 29 int sgn(double x)//符号函数，eps使用最多的地方
 30 {
 31     if (fabs(x) < eps)
 32         return 0;
 33     if (x < 0)
 34         return -1;
 35     else
 36         return 1;
 37 }
 38 
 39 //点类及其相关操作
 40 struct Point
 41 {
 42     double x, y;
 43     Point() {}
 44     Point(double _x, double _y) : x(_x), y(_y) {}
 45     Point operator-(const Point &b) const { return Point(x - b.x, y - b.y); }
 46     Point operator+(const Point &b) const { return Point(x + b.x, y + b.y); }
 47 
 48     double operator^(const Point &b) const { return x * b.y - y * b.x; } //叉积
 49     double operator*(const Point &b) const { return x * b.x + y * b.y; } //点积
 50 
 51     bool operator<(const Point &b) const { return x < b.x || (x == b.x && y < b.y); }
 52     bool operator==(const Point &b) const { return sgn(x - b.x) == 0 && sgn(y - b.y) == 0; }
 53 
 54     Point Rotate(double B, Point P) //绕着点P，逆时针旋转角度B(弧度)
 55     {
 56         Point tmp;
 57         tmp.x = (x - P.x) * cos(B) - (y - P.y) * sin(B) + P.x;
 58         tmp.y = (x - P.x) * sin(B) + (y - P.y) * cos(B) + P.y;
 59         return tmp;
 60     }
 61 };
 62 
 63 double dist(Point a, Point b) { return sqrt((a - b) * (a - b)); } //两点间距离
 64 double len(Point a){return sqrt(a.x * a.x + a.y * a.y);}//向量的长度
 65 
 66 struct Line
 67 {
 68     Point s, e;
 69     Line() {}
 70     Line(Point _s, Point _e) : s(_s), e(_e) {}
 71 
 72     //两直线相交求交点
 73     //第一个值为0表示直线重合，为1表示平行,为2是相交
 74     //只有第一个值为2时，交点才有意义
 75 
 76     pair<int, Point> operator&(const Line &b) const
 77     {
 78         Point res = s;
 79         if (sgn((s - e) ^ (b.s - b.e)) == 0)
 80         {
 81             if (sgn((s - b.e) ^ (b.s - b.e)) == 0)
 82                 return make_pair(0, res); //重合
 83             else
 84                 return make_pair(1, res); //平行
 85         }
 86         double t = ((s - b.s) ^ (b.s - b.e)) / ((s - e) ^ (b.s - b.e));
 87         res.x += (e.x - s.x) * t;
 88         res.y += (e.y - s.y) * t;
 89         return make_pair(2, res);
 90     }
 91 };
 92 
 93 //*判断点在直线上的垂点
 94 Point PointToLine(Point P, Line L)
 95 {
 96     Point result;
 97     double t = ((P - L.s) * (L.e - L.s)) / ((L.e - L.s) * (L.e - L.s));
 98     result.x = L.s.x + (L.e.x - L.s.x) * t;
 99     result.y = L.s.y + (L.e.y - L.s.y) * t;
100     return result;
101 }
102 
103 bool used[N];
104 Point points[N];
105 int n;
106 bool check(Line a)
107 {
108     //debug1((PointToLine(points[4],a) == points[4]));
109     memset(used, 0, sizeof(used));
110 
111     for(int i = 0;i < n;i ++)
112     {
113         if(PointToLine(points[i],a) == points[i])used[i] = 1;
114         else used[i] = 0;
115     }
116 
117     int cnt = 0;
118     Point t1,t2;
119     
120     for(int i = 0;i < n;i ++)if(!used[i])
121     {
122         //debug2("re",i);
123         cnt ++;
124         if(cnt == 2)
125         {
126             t2 = points[i];
127             break;
128         }
129         t1 = points[i];
130     }
131 
132     
133     if(cnt < 2)return 1;
134     else{
135         Line remainder_line(t1,t2);
136         for(int i = 0;i < n;i ++)if(!used[i])
137         {
138             if(PointToLine(points[i],remainder_line) == points[i]) used[i] = 1;
139             else return 0;
140         }
141     }
142     return 1;
143 }
144 
145 void solve() 
146 {
147     cin >> n;
148     for(int i = 0;i < n;i ++)
149     {
150         int x,y;cin >> x;cin >> y;
151         points[i].x = x; points[i].y = y;
152         //debug2(points[i].x,points[i].y);
153     }
154 
155     if(n <= 3)
156     {
157         cout << "YES" << endl;
158         return ;
159     }
160 
161     Line l0(points[0],points[1]);
162     Line l1(points[1],points[2]);
163     Line l2(points[0],points[2]);
164 
165     if(check(l0) || check(l1) || check(l2))
166     {
167         cout << "YES" << endl;
168     }else{
169         cout << "NO" << endl;
170     }
171 
172 }
173 
174 signed main()
175 {
176     
177     /*
178     ios::sync_with_stdio(false);
179     cin.tie(0);
180     cout.tie(0);
181     */
182     int T = 1;//cin >> T;
183 
184     while(T--){
185         //puts(solve()?"YES":"NO");
186         solve();
187     }
188     return 0;
189 
190 }
191 /*
192 
193 */