【POJ 3335】 Rotating Scoreboard (多边形的核- - 半平面交应用)
Rotating ScoreboardDescription
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 0Sample Output
YES NOSource
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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 }
2016-12-26 18:51:45