# poj 3335（半平面交）

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Rotating Scoreboard

 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 5158 Accepted: 2061

Description

This year, ACM/ICPC World finals will be held in a hall in form of a simple polygon. The coaches and spectators are seated along the edges of the polygon. We want to place a rotating scoreboard somewhere in the hall such that a spectator sitting anywhere on the boundary of the hall can view the scoreboard (i.e., his line of sight is not blocked by a wall). Note that if the line of sight of a spectator is tangent to the polygon boundary (either in a vertex or in an edge), he can still view the scoreboard. You may view spectator‘s seats as points along the boundary of the simple polygon, and consider the scoreboard as a point as well. Your program is given the corners of the hall (the vertices of the polygon), and must check if there is a location for the scoreboard (a point inside the polygon) such that the scoreboard can be viewed from any point on the edges of the polygon.

Input

The first number in the input line, T is the number of test cases. Each test case is specified on a single line of input in the form n x1 y1 x2 y2 ... xn yn where n (3 ≤ n ≤ 100) is the number of vertices in the polygon, and the pair of integers xi yi sequence specify the vertices of the polygon sorted in order.

Output

The output contains T lines, each corresponding to an input test case in that order. The output line contains either YES or NO depending on whether the scoreboard can be placed inside the hall conforming to the problem conditions.

Sample Input

```2
4 0 0 0 1 1 1 1 0
8 0 0  0 2  1 2  1 1  2 1  2 2  3 2  3 0
```

Sample Output

```YES
NO
```

Source

Tehran 2006 Preliminary

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``` 1 #include <stdio.h>
2 #include <string.h>
3 #include <stdlib.h>
4 #include <iostream>
5 #include <algorithm>
6 #include <math.h>
7
8 using namespace std;
9
10 #define eps 1e-8
11 #define MAXX 105
12 typedef struct
13 {
14     double x;
15     double y;
16 }point;
17
18 point p[MAXX],s[MAXX];
19
20 bool dy(double x,double y) {return x>y+eps; }
21 bool xy(double x,double y) {return x<y-eps; }
22 bool dyd(double x,double y){return x>y-eps; }
23 bool xyd(double x,double y){return x<y+eps; }
24 bool dd(double x,double y) {return fabs(x-y)<eps; }
25
26 double crossProduct(point a,point b,point c)
27 {
28     return (c.x-a.x)*(b.y-a.y)-(c.y-a.y)*(b.x-a.x);
29 }
30
31 point IntersectPoint(point u1,point u2,point v1,point v2)
32 {
33     point ans=u1;
34     double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x))/
35              ((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x));
36     ans.x += (u2.x-u1.x)*t;
37     ans.y += (u2.y-u1.y)*t;
38     return ans;
39 }
40
41 void cut(point p[],point s[],int n,int &len)
42 {
43     point tp[MAXX];
44     p[n]=p[0];
45     for(int i=0; i<=n; i++)
46     {
47         tp[i]=p[i];
48     }
49     int cp=n,tc;
50     for(int i=0; i<n; i++)
51     {
52         tc=0;
53         for(int k=0; k<cp; k++)
54         {
55             if(dyd(crossProduct(p[i],p[i+1],tp[k]),0.0))
56                 s[tc++]=tp[k];
57             if(xy(crossProduct(p[i],p[i+1],tp[k])*
58                   crossProduct(p[i],p[i+1],tp[k+1]),0.0))
59                 s[tc++]=IntersectPoint(p[i],p[i+1],tp[k],tp[k+1]);
60         }
61         s[tc]=s[0];
62         for(int k=0; k<=tc; k++)
63             tp[k]=s[k];
64         cp=tc;
65     }
66     len=cp;
67 }
68
69 int main()
70 {
71     int n,m,i,j;
72     scanf("%d",&n);
73     while(n--)
74     {
75         scanf("%d",&m);
76         for(i=0; i<m; i++)
77             scanf("%lf%lf",&p[i].x,&p[i].y);
78             int len;
79             cut(p,s,m,len);
80         if(len)printf("YES\n");
81         else printf("NO\n");
82     }
83     return 0;
84 }```

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