Problem A. Aerodynamics
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=86821#problem/A
Description
Bill is working in a secret laboratory. He is developing missiles for national security projects. Bill is the head of the aerodynamics department. One surprising fact of aerodynamics is called Whitcomb area rule. An object flying at high-subsonic speeds develops local supersonic airflows and the resulting shock waves create the effect called wave drag. Wave drag does not depend on the exact form of the object, but rather on its cross-sectional profile.
Consider a coordinate system with OZ axis pointing in the direction of object’s motion. Denote the area of a section of the object by a plane z = z0 as S(z0). Cross-sectional profile of the object is a function S that maps z0 to S(z0). There is a perfect aerodynamic shape called Sears-Haack body. The closer cross-sectional profile of an object to the cross-sectional profile of Sears-Haack body, the less wave drag it introduces. That is an essence of Whitcomb area rule.
Bill’s department makes a lot of computer simulations to study missile’s aerodynamic properties before it is even built. To approximate missile’s cross-sectional profile one takes samples of S(z0) for integer arguments z0 from zmin to zmax.
Your task is to find the area S(z0) for each integer z0 from zmin to zmax, inclusive, given the description of the missile. The description of the missile is given to you as a set of points. The missile is the minimal convex solid containing all the given points. It is guaranteed that there are four points that do not belong to the same plane.
Input
The first line of the input file contains three integer numbers: n, zmin and zmax (4 ≤ n ≤ 100, 0 ≤ zmin ≤ zmax ≤ 100). The following n lines contain three integer numbers each: x, y, and z coordinates of the given points. All coordinates do not exceed 100 by their absolute values. No two points coincide. There are four points that do not belong to the same plane.
Output
For each integer z0 from zmin to zmax, inclusive, output one floating point number: the area S(z0). The area must be precise to at least 5 digits after decimal point.
Sample Input
9 0 5
0 0 5
-3 0 2
0 -1 2
3 0 2
0 1 2
2 2 0
2 -2 0
-2 -2 0
-2 2 0
Sample Output
16.00000
14.92000
10.08000
4.48000
1.12000
0.00000
HINT
题意
给你一个由n个点构成的三维凸包，让你输出从zmin到zmax的所有截面的面积
题解：
对于每一个截面，我们n^2暴力出在这个截面上的所有点，然后直接套版求这个凸包的面积就好了
代码：
#include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define test freopen("test.txt","r",stdin) #define maxn 100010 #define mod 1000000007 #define eps 1e-9 const int inf=0x3f3f3f3f; const ll infll = 0x3f3f3f3f3f3f3f3fLL; inline ll read() { ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } //************************************************************************************** struct node { double x,y,z; }; bool cmp(node a,node b) { return a.z<b.z; } struct POINT { double x; double y; POINT(double a=0, double b=0) { x=a; y=b;} //constructor }; POINT operator - (POINT A,POINT B){return POINT(A.x-B.x,A.y-B.y);} bool cmp1(POINT a,POINT b) { if(fabs(a.x-b.x)<eps) return a.y<b.y; return a.x<b.x; } node a[110]; node c[100000]; int tot=0; POINT kiss[100000]; double Cross(POINT a,POINT b) { return a.x*b.y-a.y*b.x; } int CH(POINT* p,int n,POINT* ch) { sort(p,p+n,cmp1); int m=0; for(int i=0;i<n;i++) { while(m>1&&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&&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_of_polygon(int vcount,POINT polygon[]) { int i; double s; if(vcount<3) return 0; s=polygon[0].y*(polygon[vcount-1].x-polygon[1].x); for (i=1;i<vcount;i++) s+=polygon[i].y*(polygon[(i-1)].x-polygon[(i+1)%vcount].x); return s/2; } vector<node>Q1; vector<node>Q2; POINT ki[10010]; int main() { freopen("aerodynamics.in","r",stdin); freopen("aerodynamics.out","w",stdout); int n=read(),zmin=read(),zmax=read(); for(int i=0;i<n;i++) cin>>a[i].x>>a[i].y>>a[i].z; sort(a,a+n,cmp); int j=0; for(int i=zmin;i<=zmax;i++) { Q1.clear(); Q2.clear(); memset(kiss,0,sizeof(kiss)); memset(ki,0,sizeof(ki)); tot=0; double ii=i*1.0; while((a[j].z-ii)<-eps&&j<n) j++; for(int k=0;k<n;k++) { if(a[k].z<i) Q1.push_back((node){a[k].x,a[k].y,a[k].z}); else if(a[k].z>i) Q2.push_back((node){a[k].x,a[k].y,a[k].z}); else kiss[tot].x=a[k].x,kiss[tot++].y=a[k].y; } for(int k=0;k<Q1.size();k++) { for(int t=0;t<Q2.size();t++) { kiss[tot].x=(Q2[t].x-Q1[k].x)*(ii-Q1[k].z)/(Q2[t].z-Q1[k].z)+Q1[k].x; kiss[tot++].y=(Q2[t].y-Q1[k].y)*(ii-a[k].z)/(Q2[t].z-Q1[k].z)+Q1[k].y; } } /* if(i==4) { cout<<"--------------------------"<<endl; for(int kk=0;kk<j;kk++) cout<<a[kk].x<<" "<<a[kk].y<<" "<<a[kk].z<<endl; cout<<"--------------------------"<<endl; for(int kk=j;kk<n;kk++) cout<<a[kk].x<<" "<<a[kk].y<<" "<<a[kk].z<<endl; cout<<"--------------------------"<<endl; for(int kk=0;kk<tot;kk++) cout<<kiss[kk].x<<" "<<kiss[kk].y<<endl; cout<<"--------------------------"<<endl; } */ int ttt=CH(kiss,tot,ki); printf("%.5lf ",area_of_polygon(ttt,ki)); } }