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  • POJ 2225 / ZOJ 1438 / UVA 1438 Asteroids --三维凸包,求多面体重心

    题意: 两个凸多面体,可以任意摆放,最多贴着,问他们重心的最短距离。

    解法: 由于给出的是凸多面体,先构出两个三维凸包,再求其重心,求重心仿照求三角形重心的方式,然后再求两个多面体的重心到每个多面体的各个面的最短距离,然后最短距离相加即为答案,因为显然贴着最优。

    求三角形重心见此: http://www.cnblogs.com/whatbeg/p/4234518.html

    代码:(模板借鉴网上模板)

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    #include <cmath>
    #include <algorithm>
    #include <string>
    #include <vector>
    #include <set>
    #define Mod 1000000007
    #define eps 1e-8
    #define lll __int64
    #define ll long long
    using namespace std;
    #define N 100007
    #define MAXV 505
    
    //三维点
    struct pt{
        double x, y, z;
        pt(){}
        pt(double _x, double _y, double _z): x(_x), y(_y), z(_z){}
        pt operator - (const pt p1){return pt(x - p1.x, y - p1.y, z - p1.z);}
        pt operator * (pt p){return pt(y*p.z-z*p.y, z*p.x-x*p.z, x*p.y-y*p.x);}        //叉乘
        double operator ^ (pt p){return x*p.x+y*p.y+z*p.z;}                            //点乘
    };
    
    //pt operator - (const pt p,const pt p1){return pt(p.x - p1.x, p.y - p1.y, p.z - p1.z);}
    //pt operator ** (pt p,pt p1){return pt(p.y*p1.z-p.z*p1.y, p.z*p1.x-p.x*p1.z, p.x*p1.y-p.y*p1.x);}        //叉乘
    //double operator ^^ (pt p1,pt p){return p1.x*p.x+p1.y*p.y+p1.z*p.z;}
    
    struct _3DCH{
        struct fac{
            int a, b, c;    //表示凸包一个面上三个点的编号
            bool ok;        //表示该面是否属于最终凸包中的面
        };
    
        int n;    //初始点数
        pt P[MAXV];    //初始点
    
        int cnt;    //凸包表面的三角形数
        fac F[MAXV*8]; //凸包表面的三角形
    
        int to[MAXV][MAXV];
    
        double vlen(pt a){return sqrt(a.x*a.x+a.y*a.y+a.z*a.z);}    //向量长度
        double area(pt a, pt b, pt c){return vlen((b-a)*(c-a));}    //三角形面积*2
        double volume(pt a, pt b, pt c, pt d){return (b-a)*(c-a)^(d-a);}    //四面体有向体积*6
    
        //正:点在面同向
        double ptof(pt &p, fac &f){
            pt m = P[f.b]-P[f.a], n = P[f.c]-P[f.a], t = p-P[f.a];
            return (m * n) ^ t;
        }
        pt pvec(fac s) {
            pt k1 = (P[s.a]-P[s.b]), k2 = (P[s.b]-P[s.c]);
            return (k1*k2);
        }
        double ptoplane(pt p,fac s){
            return fabs(pvec(s)^(p-P[s.a]))/vlen(pvec(s));
        }
    
        void deal(int p, int a, int b){
            int f = to[a][b];
            fac add;
            if (F[f].ok){
                if (ptof(P[p], F[f]) > eps)
                    dfs(p, f);
                else{
                    add.a = b, add.b = a, add.c = p, add.ok = 1;
                    to[p][b] = to[a][p] = to[b][a] = cnt;
                    F[cnt++] = add;
                }
            }
        }
    
        void dfs(int p, int cur){
            F[cur].ok = 0;
            deal(p, F[cur].b, F[cur].a);
            deal(p, F[cur].c, F[cur].b);
            deal(p, F[cur].a, F[cur].c);
        }
    
        bool same(int s, int t){
            pt &a = P[F[s].a], &b = P[F[s].b], &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 construct(){
            cnt = 0;
            if (n < 4)
                return;
    
            /*********此段是为了保证前四个点不公面,若已保证,可去掉********/
            bool sb = 1;
            //使前两点不公点
            for (int i = 1; i < n; i++){
                if (vlen(P[0] - P[i]) > eps){
                    swap(P[1], P[i]);
                    sb = 0;
                    break;
                }
            }
            if (sb)return;
    
            sb = 1;
            //使前三点不公线
            for (int i = 2; i < n; i++){
                if (vlen((P[0] - P[1]) * (P[1] - P[i])) > eps){
                    swap(P[2], P[i]);
                    sb = 0;
                    break;
                }
            }
            if (sb)return;
    
            sb = 1;
            //使前四点不共面
            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]);
                    sb = 0;
                    break;
                }
            }
            if (sb)return;
            /*********此段是为了保证前四个点不公面********/
    
    
            fac add;
            for (int i = 0; i < 4; i++){
                add.a = (i+1)%4, add.b = (i+2)%4, add.c = (i+3)%4, add.ok = 1;
                if (ptof(P[i], add) > 0)
                    swap(add.b, add.c);
                to[add.a][add.b] = to[add.b][add.c] = to[add.c][add.a] = cnt;
                F[cnt++] = add;
            }
    
            for (int i = 4; i < n; i++){
                for (int j = 0; j < cnt; j++){
                    if (F[j].ok && ptof(P[i], F[j]) > eps){
                        dfs(i, j);
                        break;
                    }
                }
            }
            int tmp = cnt;
            cnt = 0;
            for (int i = 0; i < tmp; i++){
                if (F[i].ok){
                    F[cnt++] = F[i];
                }
            }
        }
    
        //表面积
        double area(){
            double ret = 0.0;
            for (int i = 0; i < cnt; i++){
                ret += area(P[F[i].a], P[F[i].b], P[F[i].c]);
            }
            return ret / 2.0;
        }
    
        //体积
        double volume(){
            pt O(0, 0, 0);
            double ret = 0.0;
            for (int i = 0; i < cnt; i++) {
                ret += volume(O, P[F[i].a], P[F[i].b], P[F[i].c]);
            }
            return fabs(ret / 6.0);
        }
    
        pt BaryCenter() {
            pt O(0, 0, 0);
            double ret = 0.0,sumvolume = 0.0, sumx = 0.0, sumy = 0.0, sumz = 0.0;
            for(int i=0;i<cnt;i++) {
                double Vol = volume(O, P[F[i].a], P[F[i].b], P[F[i].c]);
                sumvolume += Vol;
                sumx += (P[F[i].a].x + P[F[i].b].x + P[F[i].c].x)*Vol;
                sumy += (P[F[i].a].y + P[F[i].b].y + P[F[i].c].y)*Vol;
                sumz += (P[F[i].a].z + P[F[i].b].z + P[F[i].c].z)*Vol;
            }
            return pt(sumx/sumvolume/4, sumy/sumvolume/4, sumz/sumvolume/4);
        }
    
        //表面三角形数
        int facetCnt_tri(){
            return cnt;
        }
    
        //表面多边形数
        int facetCnt(){
            int ans = 0;
            for (int i = 0; i < cnt; i++){
                bool nb = 1;
                for (int j = 0; j < i; j++){
                    if (same(i, j)){
                        nb = 0;
                        break;
                    }
                }
                ans += nb;
            }
            return ans;
        }
    };
    
    
    _3DCH hull,hull2;    //内有大数组,不易放在函数内
    
    int main()
    {
        while (scanf("%d", &hull.n)!=EOF){
            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.construct();
            pt bc1 = hull.BaryCenter();
            scanf("%d",&hull2.n);
            for (int i = 0; i < hull2.n; i++)
                scanf("%lf%lf%lf", &hull2.P[i].x, &hull2.P[i].y, &hull2.P[i].z);
            hull2.construct();
            pt bc2 = hull2.BaryCenter();
            //printf("BARY1: %.2f %.2f %.2f
    ",bc1.x,bc1.y,bc1.z);
            //printf("BARY2: %.2f %.2f %.2f
    ",bc2.x,bc2.y,bc2.z);
            double dis1 = Mod, dis2 = Mod;
            for (int i = 0; i < hull.cnt; i++)
                dis1 = min(dis1,fabs(hull.ptoplane(bc1,hull.F[i])));
            for (int i = 0; i < hull2.cnt; i++)
                dis2 = min(dis2,fabs(hull2.ptoplane(bc2,hull2.F[i])));
            printf("%.6f
    ",dis1+dis2);
        }
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/whatbeg/p/4280051.html
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