五角星题解

假设逆时针时针分别为ABCDE

判断能组成五角星 等价于凸包的点还是这5个点

然后答案=S(ACE)+S(ABE)-S(ABE)

就结束了。

 1 #include<stdio.h>
 2 #include<math.h>
 3 #include<string.h>
 4 #include<iostream>
 5 #include<algorithm>
 6 #include<map>
 7 #include<set>
 8 #include<queue>
 9 #include<stack>
10 #define FOR(i,n) for(i=0;i<(n);i++)
11 #define CLR(a) memset(a,0,sizeof(a))
12 #define CIN(a) scanf("%d",&a)
13 typedef long long ll;
14 using namespace std;
15 double _x,_y;
16 double Cos(double x,double y){
17     return x/sqrt(y*y+x*x);
18 }
19 struct point{
20     double x,y;
21     point(double xx,double yy){x=xx;y=yy;}
22     point(){}
23     point operator-(const point &p){
24         return point(x-p.x,y-p.y);
25     }
26     bool operator<(const point &p) const{
27         return Cos((x-_x),(y-_y))>Cos((p.x-_x),(p.y-_y));
28     }
29     double operator*(const point &p){
30         return x*p.y-p.x*y;
31     }
32 };
33 point p[5];
34 void sw(int i,int j){
35     point x=p[i];
36     p[i]=p[j];
37     p[j]=x;
38 }
39 int pan(){
40    //printf("%lf %lf\n",(p[2]-p[1])*(p[3]-p[2]),(p[3]-p[2])*(p[4]-p[3]));
41     return (p[2]-p[1])*(p[3]-p[2])>0&&(p[3]-p[2])*(p[4]-p[3])>0;
42 }
43 double S(int i,int j,int k){
44     return fabs((p[j]-p[i])*(p[k]-p[i])/2);
45 }
46 int main()
47 {
48     int t,i;
49     //freopen("1.in","r",stdin);
50     //freopen("1.out","w",stdout);
51     scanf("%d",&t);
52     while(t--){
53         int mini=0;
54         for(i=0;i<5;i++){
55             scanf("%lf%lf",&p[i].x,&p[i].y);
56         }
57         for(i=1;i<5;i++){
58             if(p[i].y<p[mini].y) mini=i;
59         }
60         sw(i,mini);
61         _x=p[0].x;
62         _y=p[0].y;
63         sort(p+1,p+5);
64         //for(i=0;i<5;i++){
65             //printf("%lf,%lf\n",p[i].x,p[i].y);
66
67         if(pan()){
68             printf("%.5lf\n",S(0,2,4)+S(0,1,3)-S(0,1,4));
69         }else{
70             printf("-1\n");
71         }
72     }
73     return 0;
74 }