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 }
时间： 2024-12-22 10:04:08