Ultimate Weapon
| Time Limit: 2000MS | Memory Limit: 131072K | |
| Total Submissions: 1441 | Accepted: 713 |
Description
In year 2008 of the Cosmic Calendar, the Aliens send a huge armada towards the Earth seeking after conquest. The humans now depend on their ultimate weapon to retain their last hope of survival. The weapon, while capable of creating a continuous, closed and convex lethal region in the space and annihilating everything enclosed within, unfortunately exhausts upon each launch a tremendous amount of energy which is proportional to the surface area of the lethal region.
Given the positions of all battleships in the Aliens' armada, your task is to calculate the minimum amount of energy required to destroy the armada with a single launch of the ultimate weapon. You need to report the surface area of the lethal region only.
Input
The first line contains one number N -- the number of battleships.(1 ≤ N ≤ 500)
Following N lines each contains three integers presenting the position of one battleship.
Output
The minimal area rounded to three decimal places.
Sample Input
4 0 0 0 4 0 0 2 3 0 1 1 2
Sample Output
19.137
Hint
There are no four coplaner battleships.
Source
POJ Founder Monthly Contest – 2008.03.16, Jiang Liyang
模板题。
/* POJ 3528 求凸包表面积 */ #include<stdio.h> #include<algorithm> #include<string.h> #include<math.h> #include<stdlib.h> using namespace std; const int MAXN=550; const double eps=1e-8; struct Point { double x,y,z; Point(){} Point(double xx,double yy,double zz):x(xx),y(yy),z(zz){} //两向量之差 Point operator -(const Point p1) { return Point(x-p1.x,y-p1.y,z-p1.z); } //两向量之和 Point operator +(const Point p1) { return Point(x+p1.x,y+p1.y,z+p1.z); } //叉乘 Point operator *(const Point p) { return Point(y*p.z-z*p.y,z*p.x-x*p.z,x*p.y-y*p.x); } Point operator *(double d) { return Point(x*d,y*d,z*d); } Point operator / (double d) { return Point(x/d,y/d,z/d); } //点乘 double operator ^(Point p) { return (x*p.x+y*p.y+z*p.z); } }; struct CH3D { struct face { //表示凸包一个面上的三个点的编号 int a,b,c; //表示该面是否属于最终凸包上的面 bool ok; }; //初始顶点数 int n; //初始顶点 Point P[MAXN]; //凸包表面的三角形数 int num; //凸包表面的三角形 face F[8*MAXN]; //凸包表面的三角形 int g[MAXN][MAXN]; //向量长度 double vlen(Point a) { return sqrt(a.x*a.x+a.y*a.y+a.z*a.z); } //叉乘 Point cross(const Point &a,const Point &b,const 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)); } //四面体有向体积*6 double volume(Point a,Point b,Point c,Point d) { return (b-a)*(c-a)^(d-a); } //正：点在面同向 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];//搜索与该边相邻的另一个平面 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;//这里注意顺序，要成右手系 add.ok=true; g[p][b]=g[a][p]=g[b][a]=num; F[num++]=add; } } } void dfs(int p,int now)//递归搜索所有应该从凸包内删除的面 { F[now].ok=0; 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; } //构建三维凸包 void create() { int i,j,tmp; face add; num=0; if(n<4)return; //********************************************** //此段是为了保证前四个点不共面 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; //使前三个点不共线 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; //使前四个点不共面 for(int 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]; } //表面积 double area() { double res=0; if(n==3) { Point p=cross(P[0],P[1],P[2]); res=vlen(p)/2.0; return res; } for(int i=0;i<num;i++) res+=area(P[F[i].a],P[F[i].b],P[F[i].c]); return res/2.0; } double volume() { double res=0; Point tmp(0,0,0); for(int i=0;i<num;i++) res+=volume(tmp,P[F[i].a],P[F[i].b],P[F[i].c]); return fabs(res/6.0); } //表面三角形个数 int triangle() { return num; } //表面多边形个数 int polygon() { int i,j,res,flag; for(i=res=0;i<num;i++) { flag=1; for(j=0;j<i;j++) if(same(i,j)) { flag=0; break; } res+=flag; } return res; } //三维凸包重心 Point barycenter() { Point ans(0,0,0),o(0,0,0); double all=0; for(int i=0;i<num;i++) { double vol=volume(o,P[F[i].a],P[F[i].b],P[F[i].c]); ans=ans+(o+P[F[i].a]+P[F[i].b]+P[F[i].c])/4.0*vol; all+=vol; } ans=ans/all; return ans; } //点到面的距离 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; int main() { // freopen("in.txt","r",stdin); // freopen("out.txt","w",stdout); while(scanf("%d",&hull.n)==1) { for(int i=0;i<hull.n;i++) { scanf("%lf%lf%lf",&hull.P[i].x,&hull.P[i].y,&hull.P[i].z); } hull.create(); printf("%.3f\n",hull.area());//POJ 的G++用 %f提交 } return 0; }