hdu3685

Rotational Painting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 683    Accepted Submission(s): 184

Problem Description

Josh Lyman is a gifted painter. One of his great works is a glass painting. He creates some well-designed lines on one side of a thick and polygonal glass, and renders it by some special dyes. The most fantastic thing is that it can generate different meaningful paintings by rotating the glass. This method of design is called “Rotational Painting (RP)” which is created by Josh himself. 

You are a fan of Josh and you bought this glass at the astronomical sum of money. Since the glass is thick enough to put erectly on the table, you want to know in total how many ways you can put it so that you can enjoy as many as possible different paintings hiding on the glass. We assume that material of the glass is uniformly distributed. If you can put it erectly and stably in any ways on the table, you can enjoy it. 

More specifically, if the polygonal glass is like the polygon in Figure 1, you have just two ways to put it on the table, since all the other ways are not stable. However, the glass like the polygon in Figure 2 has three ways to be appreciated. 

Pay attention to the cases in Figure 3. We consider that those glasses are not stable.

 

Input

The input file contains several test cases. The first line of the file contains an integer T representing the number of test cases. 

For each test case, the first line is an integer n representing the number of lines of the polygon. (3<=n<=50000). Then n lines follow. The ith line contains two real number xi and yi representing a point of the polygon. (xi, yi) to (xi+1, yi+1) represents a edge of the polygon (1<=i<n), and (xn,yn) to (x1, y1) also represents a edge of the polygon. The input data insures that the polygon is not self-crossed.

 

Output

For each test case, output a single integer number in a line representing the number of ways to put the polygonal glass stably on the table.

 

Sample Input

  
  

   
   2
4
0 0
100 0
99 1
1 1
6
0 0
0 10
1 10
1 1
10 1
10 0
  
  

 

Sample Output

  
  

   
   2
3

   
   

    
    

     
     Hint
    
    
The sample test cases can be demonstrated by Figure 1 and Figure 2 in Description part. 
   
   
   
    
  
  

 

Source

The 35th ACM-ICPC Asia Regional Contest (Hangzhou)

 

Recommend

lcy&zhengfeng

 

 

东拼西凑模版终于被哥A掉了。。

 

#include<iostream> using namespace std; #include<cmath> #define PI acos(-1.0) const double eps=1e-6; #define zero(x) (((x)>0?(x):-(x))<eps) #define _max 50005 int sgn(double a) { return (a>eps)-(a<-eps); } struct point { double x,y; point (double xx=0,double yy=0) { //构造函数 x=xx; y=yy; } point operator -(point b) { return point(x-b.x,y-b.y); } double operator *(point b) { return (x*b.y-y*b.x); } double len() { return sqrt(x*x+y*y); } void input() { scanf("%lf%lf",&x,&y); } }p[_max]; int stack[_max],top; point Gravity(int n)// { double area = 0; point center; center.x = 0; center.y = 0; p[n] = p[0]; for(int i = 0; i < n; i++) { area += (p[i].x * p[i + 1].y - p[i].y * p[i + 1].x) / 2; // center.x += (p[i].x * p[i + 1].y - p[i].y * p[i + 1].x) * (p[i].x + p[i + 1].x); center.y += (p[i].x * p[i + 1].y - p[i].y * p[i + 1].x) * (p[i].y + p[i + 1].y); } center.x /= 6 * area; center.y /= 6 * area; return center; } int cmp(const void *a,const void *b) //逆时针排序 返回正数要交换 { struct point *c=(struct point *)a; struct point *d=(struct point *)b; double k=(*c-p[0])*(*d-p[0]); if(sgn(k)<0) return 1; else if(sgn(k)==0 && sgn((*c-p[0]).len()-(*d-p[0]).len())>=0) return 1; else return -1; } void graham(int n) //形成凸包 { int i; for(i=0;i<=2;i++) stack[i]=i; top=2; for(i=3;i<n;i++) { while( sgn((p[i]-p[stack[top-1]])*(p[stack[top]]-p[stack[top-1]])) > 0) { top--; if(top==0) break; } top++; stack[top]=i; } } point intersection(point u1,point u2,point v1,point v2) { point ret=u1; double t=((u1.x-v1.x)*(v1.y-v2.y)-(u1.y-v1.y)*(v1.x-v2.x)) /((u1.x-u2.x)*(v1.y-v2.y)-(u1.y-u2.y)*(v1.x-v2.x)); ret.x+=(u2.x-u1.x)*t; ret.y+=(u2.y-u1.y)*t; return ret; } //点到直线上的最近点 point ptoline(point p,point l1,point l2) { point t=p; t.x+=l1.y-l2.y,t.y+=l2.x-l1.x; return intersection(p,t,l1,l2); } //计算cross product (P1-P0)x(P2-P0) double xmult(point p1,point p2,point p0) { return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y); } //判点是否在线段上,包括端点 int dot_online_in(point p,point l1,point l2) { return zero(xmult(p,l1,l2))&&(l1.x-p.x)*(l2.x-p.x)<eps&&(l1.y-p.y)*(l2.y-p.y)<eps; } //判点是否在线段上,不包括端点 int dot_online_ex(point p,point l1,point l2) { return dot_online_in(p,l1,l2)&&(!zero(p.x-l1.x)||!zero(p.y-l1.y))&&(!zero(p.x-l2.x)||!zero(p.y-l2.y)); } int main() { int n,i; int cas; while(scanf("%d",&cas)!=EOF) { while(cas--) { scanf("%d",&n); for(i=0;i<n;i++) p[i].input(); point center=Gravity(n); int u=0; int cnt=0; for(i=1;i<n;i++) //找左下角的点 { if(p[i].y<p[u].y || (p[i].y==p[u].y&&p[i].x<p[u].x) ) u=i; } //swap point tmp=p[0]; p[0]=p[u]; p[u]=tmp; qsort(p+1,n-1,sizeof(p[0]),cmp); graham(n);//stack[]保存凸包 //cout<<"top "<<top<<endl; stack[top+1]=stack[0]; for(i=0;i<=top;i++) { point po=ptoline(center,p[stack[i]],p[stack[i+1]]); if(dot_online_ex(po,p[stack[i]],p[stack[i+1]])) cnt++; } cout<<cnt<<endl; } } return 0; }