HDOJ4449解题报告
题目地址：
http://acm.hdu.edu.cn/showproblem.php?pid=4449
题目概述：
在空间中给出一个凸多面体，现在让你旋转这个多面体使得它有一面与平面z=0重合，此时多面体有一个在平面z=0上的投影以及多面体上的点到平面z=0的最大距离，求旋转过程中所有这些距离中最大的值，距离相同的投影面积应该尽可能的小。
大致思路：
计算几何显然啦，首先来一个三维凸包找到这个凸多面体，然后就要开始旋转了。
小知识：把平面P：ax+by+cz=d 旋转、平移到z=0，方法是设一个向量V为向量（a,b,c）与（0,0,1）的叉积，则P绕着V做逆时针旋转，角度为前面两个向量的夹角，然后对这个凸多面体的所有点做同样的操作即可。
这样问题就只剩下求投影面积了，既然你会三维凸包，那底面显然直接上二维凸包求一个面积即可。
复杂度分析：
计算几何大部分题时间都不是大问题，代码复杂度才是最大的！！！！
代码：
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <vector> #include <ctime> #include <map> #include <assert.h> #include <stack> #include <set> #include <bitset> #include <queue> #include <iomanip> #include <cstring> #include <algorithm> using namespace std; #define sacnf scanf #define scnaf scanf #define scnafi scanfi #define pb push_back #define Len size() #define gchar getchar() #define FOR(i,j,k) for(int (i)=(j);(i)<=(k);++(i)) #define mes(a,b) memset((a),(b),sizeof(a)) #define scanfi(a) scanf("%d",&(a)) #define scanfti(a,b) scanf("%d%d",&(a),&(b)) #define println(a) printf("%d\n",(a)) #define printIs(a) printf((a)?"Yes\n":"No\n") #define print_b printf("\n") #define IsNum(x) '0'<=(x)&&(x)<='9' #define lt dir*2 #define rt dir*2+1 #define maxn 60 #define maxm 26 #define inf 1061109567 //const long long inf=1e15; #define INF 0x3f3f3f3f #define eps 1e-8 #define E 2.718281828459 const double PI=acos(-1.0); #define mod 1000 //const long long mod=1000000007; #define MAXNUM 1000000000 typedef long long ll; typedef unsigned long long ulld; ll Abs(ll x) {return (x<0)?-x:x;} int Read() {char ch;int res=0;while(ch=getchar(),!(IsNum(ch)));while(IsNum(ch)) res=res*10+ch-'0',ch=getchar();return res;} int sgn(double x) { if(x<-eps) return -1; if(x>eps) return 1; return 0; } struct vec { double x,y; vec() {x=y=0;} vec(double _x,double _y) {x=_x;y=_y;} vec operator - (vec v) {return vec(x-v.x,y-v.y);} }; double cross(vec a,vec b) {return a.x*b.y-a.y*b.x;} bool cmpXY(vec a,vec b) { if(sgn(a.x-b.x)) return a.x<b.x; return a.y<b.y; } int convex_hull(vec* v,int n,vec *z) { sort(v,v+n,cmpXY); int m=0; FOR(i,0,n-1) { while(m>1&&cross(z[m-1]-z[m-2],v[i]-z[m-2])<=0) --m; z[m++]=v[i]; } int k=m; for(int i=n-2;i>=0;--i) { while(m>k&&cross(z[m-1]-z[m-2],v[i]-z[m-2])<=0) --m; z[m++]=v[i]; } if(n>1) --m; return m; } #define PR 1e-8 struct TPoint { double x,y,z; TPoint() {} TPoint(double _x,double _y,double _z):x(_x),y(_y),z(_z) {} TPoint operator + (const TPoint p) {return TPoint(x+p.x,y+p.y,z+p.z);} TPoint operator - (const TPoint p) {return TPoint(x-p.x,y-p.y,z-p.z);} TPoint operator * (const TPoint p) {return TPoint(y*p.z-z*p.y,z*p.x-x*p.z,x*p.y-y*p.x);} //叉积 TPoint operator * (double p) {return TPoint(x*p,y*p,z*p);} TPoint operator / (double p) {return TPoint(x/p,y/p,z/p);} double operator ^ (const TPoint p) {return x*p.x+y*p.y+z*p.z;} } center; struct fac { int a,b,c; bool ok; }; struct T3dhull { int n;//初始点的个数 TPoint ply[maxn];//初始点 int trianglecnt;//凸包上三角形数 fac tri[maxn*8];//凸包三角形 int vis[maxn][maxn];//点i到点j是属于哪个面 double dist(TPoint a) {return sqrt(a.x*a.x+a.y*a.y+a.z*a.z);} double area(TPoint a,TPoint b,TPoint c) {return dist((b-a)*(c-a));}//三角形面积*2 double volume(TPoint a ,TPoint b,TPoint c,TPoint d) {return (b-a)*(c-a)^(d-a);}//四面体有向体积*6 double ptoplane(TPoint &p,fac &f) //正：点在面同向 { TPoint m=ply[f.b]-ply[f.a],n=ply[f.c]-ply[f.a],t=p-ply[f.a]; return (m*n)^t; } void deal(int p,int a,int b) { int f=vis[a][b]; fac add; if(tri[f].ok) { if(ptoplane(ply[p],tri[f])>PR) dfs(p,f); else { add.a=b,add.b=a,add.c=p,add.ok=1; vis[p][b]=vis[a][p]=vis[b][a]=trianglecnt; tri[trianglecnt++]=add; } } } void dfs(int p,int cnt) //维护凸包，如果点p在凸包外更新凸包 { tri[cnt].ok=0; deal(p,tri[cnt].b,tri[cnt].a); deal(p,tri[cnt].c,tri[cnt].b); deal(p,tri[cnt].a,tri[cnt].c); } bool same(int s,int e) //判断两个面是否为同一面 { TPoint a=ply[tri[s].a],b=ply[tri[s].b],c=ply[tri[s].c]; return fabs(volume(a,b,c,ply[tri[e].a]))<PR &&fabs(volume(a,b,c,ply[tri[e].b]))<PR &&fabs(volume(a,b,c,ply[tri[e].c]))<PR; } void construct() //构建凸包 { trianglecnt=0; if(n<4) return; bool tmp=1; FOR(i,1,n-1) //前两点不共点 { if(dist(ply[0]-ply[i])>PR) { swap(ply[1],ply[i]); tmp=0;break; } } if(tmp) return; tmp=1; FOR(i,2,n-1) //前三点不共线 { if(dist((ply[0]-ply[1])*(ply[1]-ply[i]))>PR) { swap(ply[2],ply[i]); tmp=0;break; } } if(tmp) return; tmp=1; FOR(i,3,n-1) //前四点不共面 { if(fabs((ply[0]-ply[1])*(ply[1]-ply[2])^(ply[0]-ply[i]))>PR) { swap(ply[3],ply[i]); tmp=0;break; } } if(tmp) return; fac add; FOR(i,0,3) //构建初始四面体 { add.a=(i+1)%4,add.b=(i+2)%4,add.c=(i+3)%4,add.ok=1; if(ptoplane(ply[i],add)>0) swap(add.b,add.c); vis[add.a][add.b]=vis[add.b][add.c]=vis[add.c][add.a]=trianglecnt; tri[trianglecnt++]=add; } FOR(i,4,n-1) //构建更新凸包 { FOR(j,0,trianglecnt-1) { if(tri[j].ok&&(ptoplane(ply[i],tri[j]))>PR) { dfs(i,j);break; } } } int cnt=trianglecnt; trianglecnt=0; FOR(i,0,cnt-1) if(tri[i].ok) tri[trianglecnt++]=tri[i]; } double area() //表面积 { double ret=0; FOR(i,0,trianglecnt-1) ret+=area(ply[tri[i].a],ply[tri[i].b],ply[tri[i].c]); return ret/2.0; } double volume() //体积 { TPoint p(0,0,0); double ret=0; FOR(i,0,trianglecnt-1) ret+=volume(p,ply[tri[i].a],ply[tri[i].b],ply[tri[i].c]); return fabs(ret/6); } int facetri() {return trianglecnt;} //表面三角形数 int facepolygon() //表面多边形数 { int ans=0,j,k; FOR(i,0,trianglecnt-1) { for(j=0,k=1;j<i;j++) if(same(i,j)) {k=0;break;} ans+=k; } return ans; } TPoint barycenter() //重心 { TPoint ans(0,0,0),o(0,0,0); double all=0; FOR(i,0,trianglecnt-1) { double vol=volume(o,ply[tri[i].a],ply[tri[i].b],ply[tri[i].c]); ans=ans+(o+ply[tri[i].a]+ply[tri[i].b]+ply[tri[i].c])/4.0*vol; all+=vol; } return ans/all; } double ptoface(TPoint p,int i) //点到面的距离 { return fabs(volume(ply[tri[i].a],ply[tri[i].b],ply[tri[i].c],p) /area(ply[tri[i].a],ply[tri[i].b],ply[tri[i].c])); } } a; TPoint GetPoint(TPoint st,TPoint ed,TPoint tp) { double t1=(tp-st)^(ed-st); double t2=(ed-st)^(ed-st); double t=t1/t2; return st+((ed-st)*t); } TPoint Rotate(TPoint st,TPoint ed,TPoint tp,double A) //tp以ed-st为轴旋转A度 { TPoint root=GetPoint(st,ed,tp); TPoint e=(ed-st)/a.dist(ed-st); TPoint r=tp-root;TPoint vec=e*r; /*TPoint ans=r*cos(A)+vec*sin(A)+root; printf("%.f %.f %.f\n",ans.x,ans.y,ans.z);*/ return r*cos(A)+vec*sin(A)+root; } TPoint tmp[maxn]; vec tmp1[maxn],tmp2[maxn]; int main() { //freopen("data.in","r",stdin); //freopen("std.out","w",stdout); //clock_t st=clock(); int n; while(~scanfi(n)) { if(n==0) break; int x,y,z;double ansH=0,ansS=0; FOR(i,0,n-1) { scnaf("%d%d%d",&x,&y,&z); a.ply[i]=(TPoint){x*1.0,y*1.0,z*1.0}; } a.n=n;a.construct(); FOR(i,0,a.trianglecnt-1) { FOR(j,0,n-1) tmp[j]=a.ply[j]; TPoint p1=(tmp[a.tri[i].b]-tmp[a.tri[i].a])*(tmp[a.tri[i].c]-tmp[a.tri[i].a]); TPoint e(0,0,1); TPoint vec=p1*e; double A=p1^e/a.dist(p1);A=acos(A); if(sgn(A)!=0&&sgn(A-PI)!=0) { TPoint p0(0,0,0); FOR(j,0,n-1) tmp[j]=Rotate(p0,vec,tmp[j],A); } double H=0;FOR(j,0,n-1) H=max(H,a.ptoface(a.ply[j],i)),tmp1[j].x=tmp[j].x,tmp1[j].y=tmp[j].y; sort(tmp1,tmp1+n,cmpXY);int tn=0; FOR(j,0,n-1) { int t=j+1; while(t<n&&sgn(tmp1[t].x-tmp1[j].x)==0&&sgn(tmp1[t].y-tmp1[j].y)==0) t++; tmp1[tn++]=tmp1[j];j=t-1; } int m=convex_hull(tmp1,tn,tmp2); double S=0; FOR(j,1,m-2) S+=cross(tmp2[j]-tmp2[0],tmp2[j+1]-tmp2[0]); S=fabs(S); if(sgn(ansH-H)<0) { ansH=H;ansS=S; } else if(sgn(ansH-H)==0) { if(sgn(ansS-S)>0) ansS=S; } } printf("%.3f %.3f\n",ansH,ansS/2); } //clock_t ed=clock(); //printf("\n\nTime Used : %.5lf Ms.\n",(double)(ed-st)/CLOCKS_PER_SEC); return 0; }
