HDU 4449 Building Design 第37届ACM/ICPC 金华赛区H题 （计算几何，三维凸包+空间坐标旋转+二维凸包）
Building Design
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 431 Accepted Submission(s): 117
Problem Description
You are a notable architect.
Recently, a company invites you to design their new building for them. It is not an easy task, because they make some strange claims to the new building:
1.Its shape should be the same as a convex polyhedron which they have given to you, though you can rotate it arbitrarily.
2.One face of the building should cling to the ground. (Of course! Have you ever seen a building only touch the ground by an edge, or a point, or even suspend in the air?)
3.They want the building to be as striking as possible. We call the highest point of the building as the “peak”. The higher the peak is the more striking the building will be.
4.If there are many designs meet all claims above, the occupied land of the building should be less. In another word, the building's vertical projection area on the ground should be as small as possible.
Now, give you the relative positions of vertices of the building, please design the building and tell them H – the height of the peak, as well as S - the projection area of the building in your best design.
Recently, a company invites you to design their new building for them. It is not an easy task, because they make some strange claims to the new building:
1.Its shape should be the same as a convex polyhedron which they have given to you, though you can rotate it arbitrarily.
2.One face of the building should cling to the ground. (Of course! Have you ever seen a building only touch the ground by an edge, or a point, or even suspend in the air?)
3.They want the building to be as striking as possible. We call the highest point of the building as the “peak”. The higher the peak is the more striking the building will be.
4.If there are many designs meet all claims above, the occupied land of the building should be less. In another word, the building's vertical projection area on the ground should be as small as possible.
Now, give you the relative positions of vertices of the building, please design the building and tell them H – the height of the peak, as well as S - the projection area of the building in your best design.
Input
There are several test cases in the input.
Each test case begins with one integer n (1 ≤ n ≤ 50), indicating the number of vertices of the building.
Then n lines follow. Each line contains three integers x, y, z (-10000 ≤ x, y, z ≤ 10000), separated by spaces, indicating the relative positions of one vertex of the building.
The input ends with n = 0.
Each test case begins with one integer n (1 ≤ n ≤ 50), indicating the number of vertices of the building.
Then n lines follow. Each line contains three integers x, y, z (-10000 ≤ x, y, z ≤ 10000), separated by spaces, indicating the relative positions of one vertex of the building.
The input ends with n = 0.
Output
For each test case, output two float numbers H and S in one line, separated by one space.
Please round the results to three digits after decimal point.
Please round the results to three digits after decimal point.
Sample Input
6 1 0 0 -1 0 0 0 1 0 0 -1 0 0 0 1 0 0 -1 0
Sample Output
1.155 1.732
Source
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zhuyuanchen520
题目就是给了一个凸多面题。
要把这个凸多面体放到地上，一个面要接触到地面。
求一种放法，使得最高点最高，最高点相同时使投影面积最小。输出最高点高度H和投影面积S。
先三维凸包一下求得所有的三角形表面。
然后枚举每一个表面，进行坐标变换，旋转使得这个表面变水平，然后投影求投影面积，算最高点高度。
题目的数据好像没有n==1 ,2,3的情况，可以忽略。
有个问题就是如果按照旋转之后的z坐标值来算高度的话，会产生很大的精度差。最好就是直接算点到面的距离。
//============================================================================ // Name : Building.cpp // Author : kuangbin // Version : // Copyright : Your copyright notice // Description : HDU 4449 // 三维凸包+二维凸包+三维坐标旋转变换 //============================================================================ #include <iostream> #include <string.h> #include <math.h> #include <algorithm> #include <stdio.h> #include <vector> #include <map> #include <set> #include <queue> using namespace std; const int MAXN = 110; const double eps=1e-8; const double PI=acos(-1.0); inline int sign(double p) { if(p>eps)return 1; else if(p<-eps)return -1; else return 0; } struct Point { double x,y,z; Point(double xx=0,double yy=0,double zz=0):x(xx),y(yy),z(zz){} Point operator +(const Point p1) { return Point(x+p1.x,y+p1.y,z+p1.z); } Point operator -(const Point p1) { return Point(x-p1.x,y-p1.y,z-p1.z); } Point operator *(const Point p1) { return Point(y*p1.z-z*p1.y,z*p1.x-x*p1.z,x*p1.y-y*p1.x); } Point operator *(double d) { return Point(d*x,d*y,d*z); } Point operator /(double d) { return Point(x/d,y/d,z/d); } double operator ^(const Point p1) { return x*p1.x+y*p1.y+z*p1.z; } double getLen() { return sqrt(x*x+y*y+z*z); } void read() { scanf("%lf%lf%lf",&x,&y,&z); } }; struct face { int a,b,c; bool ok; }; struct CH3D { int n;//初始顶点数 Point P[MAXN];//初始顶点 int num;//凸包表面的三角形个数 face F[8*MAXN];//凸包表面的三角形 int g[MAXN][MAXN]; double vlen(Point a) { return sqrt(a.x*a.x+a.y*a.y+a.z*a.z); } Point cross(Point a,Point b,Point c) { return Point( (b.y-a.y)*(c.z-a.z)-(b.z-a.z)*(c.y-a.y), (b.z-a.z)*(c.x-a.x)-(b.x-a.x)*(c.z-a.z), (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x) ); } //三角形面积*2 double area(Point a,Point b,Point c) { return vlen((b-a)*(c-a)); } //四面体面积*6 double volume(Point a,Point b,Point c,Point d) { return ((b-a)*(c-a))^(d-a); } double dblcmp(Point &p,face &f) { Point m=P[f.b]-P[f.a]; Point n=P[f.c]-P[f.a]; Point t=p-P[f.a]; return (m*n)^t; } void deal(int p,int a,int b) { int f=g[a][b]; face add; if(F[f].ok) { if(dblcmp(P[p],F[f])>eps) dfs(p,f); else { add.a=b; add.b=a; add.c=p; add.ok=true; g[p][b]=g[a][p]=g[b][a]=num; F[num++]=add; } } } void dfs(int p,int now) { F[now].ok=false; deal(p,F[now].b,F[now].a); deal(p,F[now].c,F[now].b); deal(p,F[now].a,F[now].c); } bool same(int s,int t) { Point &a=P[F[s].a]; Point &b=P[F[s].b]; Point &c=P[F[s].c]; return fabs(volume(a,b,c,P[F[t].a]))<eps && fabs(volume(a,b,c,P[F[t].b]))<eps && fabs(volume(a,b,c,P[F[t].c]))<eps; } void create() { int i,j,tmp; face add; num=0; if(n<4)return; bool flag=true; for(i=1;i<n;i++) { if(vlen(P[0]-P[i])>eps) { swap(P[1],P[i]); flag=false; break; } } if(flag)return; flag=true; for(i=2;i<n;i++) { if(vlen((P[0]-P[1])*(P[1]-P[i]))>eps) { swap(P[2],P[i]); flag=false; break; } } if(flag)return; flag=true; for(i=3;i<n;i++) { if(fabs((P[0]-P[1])*(P[1]-P[2])^(P[0]-P[i]))>eps) { swap(P[3],P[i]); flag=false; break; } } if(flag)return; for(i=0;i<4;i++) { add.a=(i+1)%4; add.b=(i+2)%4; add.c=(i+3)%4; add.ok=true; if(dblcmp(P[i],add)>0)swap(add.b,add.c); g[add.a][add.b]=g[add.b][add.c]=g[add.c][add.a]=num; F[num++]=add; } for(i=4;i<n;i++) { for(j=0;j<num;j++) { if(F[j].ok && dblcmp(P[i],F[j])>eps) { dfs(i,j); break; } } } tmp=num; for(i=num=0;i<tmp;i++) if(F[i].ok) F[num++]=F[i]; } double ptoface(Point p,int i) { return fabs(volume(P[F[i].a],P[F[i].b],P[F[i].c],p)/vlen((P[F[i].b]-P[F[i].a])*(P[F[i].c]-P[F[i].a]))); } }; CH3D hull; Point ps[MAXN]; int n; inline Point get_point(Point st,Point ed,Point tp) { double t1=(tp-st)^(ed-st); double t2=(ed-st)^(ed-st); double t=t1/t2; Point ans=st + ((ed-st)*t); return ans; } inline double dist(Point st,Point ed) { return sqrt((ed-st)^(ed-st)); } Point rotate(Point st,Point ed,Point tp,double A) { Point root=get_point(st,ed,tp); Point e=(ed-st)/dist(ed,st); Point r=tp-root; Point vec=e*r; Point ans=r*cos(A)+vec*sin(A)+root; return ans; } struct point { double x,y; }; point list[MAXN]; int stack[MAXN],top; double cross(point p0,point p1,point p2) { return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x); } double dis(point p1,point p2) { return sqrt((p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y)); } bool cmp(point p1,point p2) { double tmp=cross(list[0],p1,p2); if(sign(tmp)>0)return true; else if(sign(tmp)==0 && dis(list[0],p1)<dis(list[0],p2))return true; else return false; } void init(int n) { point p0; p0=list[0]; int k=0; for(int i=1;i<n;i++) { if( (p0.y+eps > list[i].y) || ( sign(p0.y-list[i].y)==0 && (p0.x+eps > list[i].x))) { p0=list[i]; k=i; } } list[k]=list[0]; list[0]=p0; sort(list+1,list+n,cmp); } void graham(int n) { init(n); if(n==1) { top=0; stack[0]=0; return; } if(n==2) { top=1; stack[0]=0; stack[1]=1; return; } for(int i=0;i<=1;i++)stack[i]=i; top=1; for(int i=2;i<n;i++) { while(top>0 && cross(list[stack[top-1]],list[stack[top]],list[i])<=eps) top--; top++; stack[top]=i; } } bool input() { scanf("%d",&n); if(n==0)return false; hull.n=n; for(int i=0;i<n;i++) hull.P[i].read(); return true; } void solve() { double ansH=0,ansS=1e200; double H,S; if(n<=2)//该题数据好像没有这种情况 { printf("0.000 0.000\n"); return; } if(n==3)//该题数据好像没有这种情况 { ansH=0; ansS=(hull.P[1]-hull.P[0])^(hull.P[2]-hull.P[0]); ansS/=2.0; printf("%.3lf %.3lf\n",ansH,ansS); return; } hull.create(); for(int i=0;i< hull.num;i++) { for(int j=0;j<n;j++)ps[j]=hull.P[j]; Point p1=(ps[hull.F[i].b]-ps[hull.F[i].a])*(ps[hull.F[i].c]-ps[hull.F[i].a]); Point e(0,0,1); Point vec=p1*e; double A=p1^e/p1.getLen(); A=acos(A); if(sign(A)!=0 && sign(A-PI)!=0) { Point p0(0,0,0); for(int j=0;j<n;j++) ps[j]=rotate(p0,vec,ps[j],A); } double tt=ps[hull.F[i].a].z; for(int j=0;j<n;j++) ps[j].z -= tt; H=0; for(int j=0;j<n;j++) { //H=max(H,fabs(ps[j].z)+0.00001);//这个有精度差，一直会WA H=max(H,hull.ptoface(hull.P[j],i)); //if( fabs(hull.ptoface(hull.P[j],i)-fabs(ps[j].z))>=0.00000005)while(1); //加了上面这句不超时，说明两者很接近了，由于精度问题会一直WA } //for(int j=0;j<n;j++) // H=max(H,hull.ptoface(hull.P[j],i)); for(int j=0;j<n;j++) { list[j].x=ps[j].x; list[j].y=ps[j].y; } graham(n); S=0; for(int j=0;j<top;j++) S+=(list[stack[j]].x*list[stack[j+1]].y - list[stack[j]].y*list[stack[j+1]].x); S+=(list[stack[top]].x*list[stack[0]].y - list[stack[top]].y*list[stack[0]].x); S=fabs(S); S/=2; if(H > ansH || ( sign(H-ansH)==0 && ansS > S )) { ansH=H; ansS=S; } } printf("%.3lf %.3lf\n",ansH,ansS); } int main() { while(input())solve(); return 0; }