[SCOI2007]最大土地面积

Posted 2020-10-20 Z-Y-Y-S

Description

　　在某块平面土地上有N个点，你可以选择其中的任意四个点，将这片土地围起来，当然，你希望这四个点围成
的多边形面积最大。

Input

　　第1行一个正整数N，接下来N行，每行2个数x,y，表示该点的横坐标和纵坐标。

Output

　　最大的多边形面积，答案精确到小数点后3位。

Sample Input

5
0 0
1 0
1 1
0 1
0.5 0.5

Sample Output

1.000

HINT

数据范围 n<=2000, |x|,|y|<=100000

首先最大的4个点一定在凸包上

先把凸包求出来

然后选定对角线，然后发现另外两个顶点是单调的，用旋转卡壳的思想

 

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cmath>
 6 using namespace std;
 7 struct point
 8 {
 9   double x,y;
10 }p[5001],s[5001];
11 int n,top;
12 double ans,eps=1e-5;
13 int dcmp(double x)
14 {
15   if (x<-eps) return -1;
16   if (x>eps) return 1;
17   return 0;
18 }
19 double operator *(point a,point b)
20 {
21   return a.x*b.y-a.y*b.x;
22 }
23 point operator -(point a,point b)
24 {
25   return (point){a.x-b.x,a.y-b.y};
26 }
27 double dist(point a,point b)
28 {
29   return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
30 }
31 bool cmp(point a,point b)
32 {
33   return (a.y<b.y)||(a.y==b.y&&a.x<b.x);
34 }
35 bool cmp2(point a,point b)
36 {
37   int t=dcmp((p[1]-a)*(p[1]-b));
38   if (t==0) return dist(p[1],a)<dist(p[1],b);
39   return t>0;
40 }
41 void graham()
42 {int i;
43   sort(p+1,p+n+1,cmp);
44   sort(p+2,p+n+1,cmp2);
45   s[++top]=p[1];s[++top]=p[2];
46   for (i=3;i<=n;i++)
47     {
48       while (top>1&&dcmp((p[i]-s[top-1])*(s[top]-s[top-1]))>=0) top--;
49       s[++top]=p[i];
50     }
51 }
52 void solve()
53 {int i,j;
54   s[top+1]=s[1];
55   for (i=1;i<=top-2;i++)
56     {
57       int a=i%top+1,b=(i+2)%top+1;
58       for (j=i+2;j<=top;j++)
59     {
60       while (a%top+1!=j&&dcmp((s[a+1]-s[i])*(s[j]-s[i])-(s[a]-s[i])*(s[j]-s[i]))>0) a=a%top+1;
61       while (b%top+1!=i&&dcmp((s[j]-s[i])*(s[b+1]-s[i])-(s[j]-s[i])*(s[b]-s[i]))>0) b=b%top+1;
62       //cout<<i<<‘ ‘<<a<<‘ ‘<<j<<‘ ‘<<b<<endl;
63       ans=max(ans,(s[a]-s[i])*(s[j]-s[i])+(s[j]-s[i])*(s[b]-s[i]));
64     }
65     }
66 }
67 int main()
68 {int i;
69   cin>>n;
70   for (i=1;i<=n;i++)
71     {
72       scanf("%lf%lf",&p[i].x,&p[i].y);
73     }
74   graham();
75   solve();
76   printf("%.3lf\n",ans/2.0);
77 }