第二周 9.5 --- 9.11

Posted 2020-08-06

新的一周>.<

9.5

cf 340 b B - Maximal Area Quadrilateral

给出 n 个点，求能够形成的四边形的最大面积

最开始的做法是 暴力枚举4个点，再去算面积，不过不对..应该是算面积那里不对

然后应该是枚举四边形的对角线

维护对角线两边的三角形的最大值

  1 #include <cstdio>
  2 #include <algorithm>
  3 #include <cmath>
  4 #include <vector>
  5 using namespace std;
  6 
  7 const int INF = (1<<30)-1;
  8 
  9 //lrj计算几何模板
 10 struct Point
 11 {
 12     int x, y;
 13     Point(double x=0, double y=0) :x(x),y(y) {}
 14 };
 15 typedef Point Vector;
 16 
 17 Point read_point(void)
 18 {
 19     double x, y;
 20     scanf("%lf%lf", &x, &y);
 21     return Point(x, y);
 22 }
 23 
 24 const double EPS = 1e-10;
 25 
 26 //向量+向量=向量 点+向量=点
 27 Vector operator + (Vector A, Vector B)    { return Vector(A.x + B.x, A.y + B.y); }
 28 
 29 //向量-向量=向量 点-点=向量
 30 Vector operator - (Vector A, Vector B)    { return Vector(A.x - B.x, A.y - B.y); }
 31 
 32 //向量*数=向量
 33 Vector operator * (Vector A, double p)    { return Vector(A.x*p, A.y*p); }
 34 
 35 //向量/数=向量
 36 Vector operator / (Vector A, double p)    { return Vector(A.x/p, A.y/p); }
 37 
 38 bool operator < (const Point& a, const Point& b)
 39 { return a.x < b.x || (a.x == b.x && a.y < b.y); }
 40 
 41 int dcmp(double x)
 42 { if(fabs(x) < EPS) return 0; else return x < 0 ? -1 : 1; }
 43 
 44 bool operator == (const Point& a, const Point& b)
 45 { return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0; }
 46 
 47 /**********************基本运算**********************/
 48 
 49 //点积
 50 double Dot(Vector A, Vector B)
 51 { return A.x*B.x + A.y*B.y; }
 52 //向量的模
 53 double Length(Vector A)    { return sqrt(Dot(A, A)); }
 54 
 55 //向量的夹角，返回值为弧度
 56 double Angle(Vector A, Vector B)
 57 { return acos(Dot(A, B) / Length(A) / Length(B)); }
 58 
 59 //叉积
 60 double Cross(Vector A, Vector B)
 61 { return A.x*B.y - A.y*B.x; }
 62 
 63 //向量AB叉乘AC的有向面积
 64 double Area2(Point A, Point B, Point C)
 65 { return Cross(B-A, C-A); }
 66 
 67 //向量A旋转rad弧度
 68 Vector VRotate(Vector A, double rad)
 69 {
 70     return Vector(A.x*cos(rad) - A.y*sin(rad), A.x*sin(rad) + A.y*cos(rad));
 71 }
 72 
 73 //将B点绕A点旋转rad弧度
 74 Point PRotate(Point A, Point B, double rad)
 75 {
 76     return A + VRotate(B-A, rad);
 77 }
 78 
 79 //求向量A向左旋转90°的单位法向量,调用前确保A不是零向量
 80 Vector Normal(Vector A)
 81 {
 82     double l = Length(A);
 83     return Vector(-A.y/l, A.x/l);
 84 }
 85 
 86 /**********************点和直线**********************/
 87 
 88 //求直线P + tv 和 Q + tw的交点，调用前要确保两条直线有唯一交点
 89 Point GetLineIntersection(Point P, Vector v, Point Q, Vector w)
 90 {
 91     Vector u = P - Q;
 92     double t = Cross(w, u) / Cross(v, w);
 93     return P + v*t;
 94 }//在精度要求极高的情况下，可以自定义分数类
 95 
 96 //P点到直线AB的距离
 97 double DistanceToLine(Point P, Point A, Point B)
 98 {
 99     Vector v1 = B - A, v2 = P - A;
100     return fabs(Cross(v1, v2)) / Length(v1);    //不加绝对值是有向距离
101 }
102 
103 //点到线段的距离
104 double DistanceToSegment(Point P, Point A, Point B)
105 {
106     if(A == B)    return Length(P - A);
107     Vector v1 = B - A, v2 = P - A, v3 = P - B;
108     if(dcmp(Dot(v1, v2)) < 0)    return Length(v2);
109     else if(dcmp(Dot(v1, v3)) > 0)    return Length(v3);
110     else return fabs(Cross(v1, v2)) / Length(v1);
111 }
112 
113 //点在直线上的射影
114 Point GetLineProjection(Point P, Point A, Point B)
115 {
116     Vector v = B - A;
117     return A + v * (Dot(v, P - A) / Dot(v, v));
118 }
119 
120 //线段“规范”相交判定
121 bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2)
122 {
123     double c1 = Cross(a2-a1, b1-a1), c2 = Cross(a2-a1, b2-a1);
124     double c3 = Cross(b2-b1, a1-b1), c4 = Cross(b2-b1, a2-b1);
125     return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
126 }
127 
128 //判断点是否在线段上
129 bool OnSegment(Point P, Point a1, Point a2)
130 {
131     Vector v1 = a1 - P, v2 = a2 - P;
132     return dcmp(Cross(v1, v2)) == 0 && dcmp(Dot(v1, v2)) < 0;
133 }
134 
135 //求多边形面积
136 double PolygonArea(Point* P, int n)
137 {
138     double ans = 0.0;
139     for(int i = 1; i < n - 1; ++i){
140         double tmp = Cross(P[i]-P[0], P[i+1]-P[0]);
141         tmp = fabs(tmp);
142         //printf("tmp = %lf\\n",tmp);
143         ans += tmp;
144     }
145     return ans/2;
146 }
147 
148 Point p[5],a[505];
149 int n;
150 
151 void solve(){
152     double ans = 0.0;
153     for(int i = 1;i <= n;i++){
154         for(int j = i+1;j <= n;j++){
155             double minn = -1.0*INF;
156             double maxx = -1.0*INF;
157             for(int k = 1;k <= n;k++){
158                 if(k == i || k == j) continue;
159 //                printf("i = %d j = %d k = %d ",i,j,k);
160                 double tmp = Cross(a[i]-a[j], a[k]-a[j]);
161                 if(tmp < 0) {
162                     tmp = -tmp;
163                     minn = max(tmp,minn);
164                 }
165                 else maxx = max(maxx,tmp);
166             }
167             ans = max(ans,(minn+maxx)/2.0);
168         }
169     }
170     printf("%.12lf\\n",ans);
171 }
172 
173 int main(){
174     while(scanf("%d",&n) != EOF){
175         for(int i = 1;i <= n;i++){
176             scanf("%d %d",&a[i].x,&a[i].y);
177         }
178         solve();
179     }
180     return 0;
181 }

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