UVa 1433 - Aerodynamics

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这是我第一次使用多个namespace 写程序，对于3D和2D要同时出现的题目完全可以把3D中的每一个变量后加一个3 ，但是完全没有下面的写法方便。

训练指南的三维凸包写法非常实用，进行些许扰动，可以减少很多麻烦。这个题指明了： no two points coincide , 所以不需要判重

提示：
1. 求凸包
2. 然后关于z的一个平面一个平面的枚举就好了

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <string>#include <vector>#include <deque>#include <stack>#include <algorithm>using namespace std;const double eps = 1e-10;const int maxn = 110;int dcmp(double a) { return fabs(a)<eps?0:(a<0?-1:1); }namespace TwoD{    struct points    {        double x , y;        void read(){ scanf("%lf%lf",&x,&y); }        points(double x = 0 , double y = 0):x(x),y(y){}        bool operator <(const points& b)const { return dcmp(x-b.x)==-1 || (dcmp(x-b.x)==0 && dcmp(y-b.y)==-1); }        bool operator ==(const points& b)const { return dcmp(x-b.x)==0 && dcmp(y-b.y)==0; }    };    typedef points Vector;    typedef vector<points> polygon;    Vector operator +(Vector a , Vector b) { return Vector(a.x+b.x , a.y+b.y); }    Vector operator -(Vector a , Vector b) { return Vector(a.x-b.x , a.y-b.y); }    Vector operator *(Vector a , double b) { return Vector(a.x*b   , a.y*b  ); }    Vector operator /(Vector a , double b) { return Vector(a.x/b   , a.y/b  ); }    double Cross(Vector a , Vector b) { return a.x*b.y-a.y*b.x; }    double Dot(Vector a , Vector b) { return a.x*b.x+a.y*b.y; }    points ch[maxn];    int convexHull(polygon p)    {        sort(p.begin(), p.end());        p.erase(unique(p.begin(), p.end()), p.end());        int n = p.size();        int m = 0;        for(int i=0;i<n;i++)        {            while(m>1 && dcmp(Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]))<=0) m--;            ch[m++] = p[i];        }        int k = m;        for(int i=n-2;i>=0;i--)        {            while(m>k && dcmp(Cross(ch[m-1]-ch[m-2], p[i]-ch[m-2]))<=0) m--;            ch[m++] = p[i];        }        if(n>1) m--;        return m;    }    double area(int m)    {        double res = 0;        for(int i=1;i<m-1;i++) res+= Cross(ch[i]-ch[0], ch[i+1]-ch[0]);        return res/2.0;    }};namespace ThreeD{    struct points    {        double x , y , z;        void read(){ scanf("%lf%lf%lf",&x,&y,&z); }        points(double x =0  , double y =0  , double z =0 ):x(x),y(y),z(z){}    };    typedef points Vector;    Vector operator +(Vector a , Vector b) { return Vector(a.x+b.x , a.y+b.y , a.z+b.z); }    Vector operator -(Vector a , Vector b) { return Vector(a.x-b.x , a.y-b.y , a.z-b.z); }    Vector operator *(Vector a , double b) { return Vector(a.x*b   , a.y*b   , a.z*b  ); }    Vector operator /(Vector a , double b) { return Vector(a.x/b   , a.y/b   , a.z/b  ); }    double Dot(Vector a , Vector b) { return a.x*b.x+a.y*b.y+a.z*b.z; }    Vector Cross(Vector a , Vector b) { return Vector(a.y*b.z-a.z*b.y , a.z*b.x-a.x*b.z , a.x*b.y-a.y*b.x); }    double rand01() { return rand() / (double)RAND_MAX; }    double randeps() { return (rand01()-0.5)*eps; }    points addnoise(points p) { return points(p.x+randeps() , p.y+randeps() , p.z+randeps()); }    struct face    {        int v[3];        face(int v1=0, int v2=0, int v3=0){ v[0] = v1; v[1] = v2; v[2] = v3; }        Vector normal(points *p)        {            return Cross(p[v[1]]-p[v[0]], p[v[2]]-p[v[0]]);        }        int cansee(points* p , int i)         {            return Dot(p[i]-p[v[0]] , normal(p))>0?1:0;        }    };    points p[maxn];    points t[maxn];    int vis[maxn][maxn];    vector<face> convexHull(points* p , int n)    {        vector<face> res;        res.push_back(face(0,1,2));        res.push_back(face(2,1,0));        for(int i=3;i<n;i++)        {            vector<face> Next;            for(int j=0;j<res.size();j++)            {                int hi = res[j].cansee(p, i);                if(!hi) Next.push_back(res[j]);                for(int k=0;k<3;k++) vis[res[j].v[k]][res[j].v[(k+1)%3]] = hi;            }            for(int j=0;j<res.size();j++) for(int k=0;k<3;k++)            {                int a = res[j].v[k] , b = res[j].v[(k+1)%3];                if(vis[a][b]!=vis[b][a] && vis[a][b]) Next.push_back(face(a,b,i));            }            res = Next;         }        return res;     }};using namespace ThreeD;int main(int argc, char *argv[]) {    int n , zmin , zmax , Case=0;    while(cin>>n>>zmin>>zmax)    {        if(Case) cout<<endl; Case++;        for(int i=0;i<n;i++) p[i].read();        for(int i=0;i<n;i++) t[i] = addnoise(p[i]);        vector<face> c = convexHull(t, n);        for(int z=zmin;z<=zmax;z++)        {            TwoD::polygon poly;            for(int i=0;i<c.size();i++) for(int j=0;j<3;j++)            {                int a = c[i].v[j] , b = c[i].v[(j+1)%3];                double z1 = p[a].z , z2 = p[b].z , x1 = p[a].x , x2 = p[b].x , y1 = p[a].y , y2 = p[b].y;                if(dcmp(z1-z2)==0) continue;                if(dcmp(p[a].z-z)*dcmp(p[b].z-z)<=0)                     poly.push_back(TwoD::points(x1+(z-z1)/(z2-z1)*(x2-x1) , y1+(z-z1)/(z2-z1)*(y2-y1)));            }            int m = TwoD::convexHull(poly);//          cout<<poly.size()<<" "<<m<<endl;             printf("%.5lf\n",TwoD::area(m));         }    }    return 0;}

程序的时间表现相当优秀:-)

详解：
1. 在枚举每一个平面的时候，找到这个平面在凸包上的所有交点，然后把这些交点排序，这里我用了二维凸包，这更加鲁棒

注意：
两个z值相等的线段不做考量，那会不会漏掉些点呢？不会，因为每个点会在多个平面中出现，所以不用担心

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