【POJ 3335】 Rotating Scoreboard （多边形的核- - 半平面交应用）

Rotating Scoreboard

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 nx1y1x2y2 ... xnyn where n (3 ≤ n ≤ 100) is the number of vertices in the polygon, and the pair of integers xiyi 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
 
 
【分析】
　　一个比较好的多边形的核的求法的解答：http://www.cnblogs.com/ka200812/archive/2012/01/20/2328316.html
 
　　把原题转换成半平面的交，看看是否为空集。
　　可以参考一下这个图：
　　
圈起来的是原多边形的顶点，其他是辅助线和辅助点，阴影部分是多边形的核。
 
  1 #include<cstdio>
  2 #include<cstdlib>
  3 #include<cstring>
  4 #include<iostream>
  5 #include<algorithm>
  6 #include<cmath>
  7 using namespace std;
  8 #define Maxn 100010
  9 
 10 const double eps=0.0001;
 11 const double pi=3.141592653;
 12 
 13 struct P {double x,y;};
 14 struct L {P a,b;double slop;}l[Maxn],p[Maxn];
 15 int len;
 16 
 17 double Dot(P x,P y) {return x.x*y.x+x.y*y.y;}
 18 double Cross(P x,P y) {return x.x*y.y-x.y*y.x;}
 19 
 20 P operator - (P x,P y)
 21 {
 22     P tt;
 23     tt.x=x.x-y.x;
 24     tt.y=x.y-y.y;
 25     return tt;
 26 }
 27 
 28 P operator + (P x,P y)
 29 {
 30     P tt;
 31     tt.x=x.x+y.x;
 32     tt.y=x.y+y.y;
 33     return tt;
 34 }
 35 
 36 P operator * (P x,double y)
 37 {
 38     P tt;
 39     tt.x=x.x*y;
 40     tt.y=x.y*y;
 41     return tt;
 42 }
 43 
 44 bool operator < (L x,L y) {return (x.slop==y.slop)?(Cross(x.b-x.a,y.b-x.a)<0):(x.slop<y.slop);}
 45 
 46 P inter(L x,L y)
 47 {
 48     P nw=y.a-x.a;
 49     P X=x.b-x.a,Y=y.b-y.a;
 50     double tt;
 51     tt=Cross(nw,X)/Cross(X,Y);
 52     return y.a+Y*tt;
 53 }
 54 
 55 bool jud(L x,L y,L z)
 56 {
 57     // if(x.slop==-y.slop) return 1;
 58     // if(x.slop-pi-y.slop<=eps&&x.slop-pi-y.slop>=-eps) return 1;
 59     // if(y.slop-pi-x.slop<=eps&&y.slop-pi-x.slop>=-eps) return 1;
 60     //???????
 61     P nw=inter(x,y);
 62     return Cross(z.b-z.a,nw-z.a)<0;
 63 }
 64 
 65 int cnt;
 66 
 67 void op()
 68 {
 69     for(int i=1;i<=cnt;i++)
 70     {
 71         printf("%.2lf %.2lf %.2lf %.2lf = %.2lf\n",l[i].a.x,l[i].a.y,l[i].b.x,l[i].b.y,l[i].slop);
 72     }
 73     printf("\n");
 74 }
 75 
 76 void opp(int L,int R)
 77 {
 78     for(int i=L;i<=R;i++)
 79     {
 80         printf("%.2lf %.2lf %.2lf %.2lf = %.2lf\n",p[i].a.x,p[i].a.y,p[i].b.x,p[i].b.y,p[i].slop);
 81     }
 82     printf("\n");
 83 }
 84 
 85 void ffind()
 86 {
 87     sort(l+1,l+1+cnt);
 88     // op();
 89     int tot=1;
 90     for(int i=2;i<=cnt;i++)
 91     {
 92         if(l[i].slop!=l[tot].slop) l[++tot]=l[i];
 93     }
 94     cnt=tot;
 95     if(cnt<=2) {printf("NO\n");return;}
 96     // op();
 97     p[1]=l[1];p[2]=l[2];
 98     int L=1,R=2;
 99     for(int i=3;i<=cnt;i++)
100     {
101         while(R>L&&jud(p[R],p[R-1],l[i])) R--;
102         while(R>L&&jud(p[L],p[L-1],l[i])) L++; 
103         p[++R]=l[i];
104     }
105     // if(R>L&&jud(p[L],p[L+1],p[R])<0) R--;
106     if(L<R&&jud(p[R-1],p[R],p[L])) R--;
107     // opp(L,R);
108     if(R-L+1<=2) printf("NO\n");
109     else printf("YES\n");
110 }
111 
112 int main()
113 {
114     int T;
115     scanf("%d",&T);
116     while(T--)
117     {
118         int n;
119         scanf("%d",&n);
120         P now,ft;
121         scanf("%lf%lf",&ft.x,&ft.y);
122         now=ft;
123         cnt=0;
124         for(int i=2;i<=n;i++)
125         {
126             P nw;
127             scanf("%lf%lf",&nw.x,&nw.y);
128             l[++cnt].a=nw;l[cnt].b=now;
129             now=nw;
130         }
131         l[++cnt].a=ft;l[cnt].b=now;
132         for(int i=1;i<=cnt;i++) l[i].slop=atan2(l[i].b.x-l[i].a.x,l[i].b.y-l[i].a.y);
133         // op();
134         ffind();
135     }
136     return 0;
137 }
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2016-12-26 18:51:45

posted @ 2016-12-26 18:46  konjak魔芋  阅读(245)  评论(0)  编辑  收藏  举报