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分类专栏：图形学|计算几何

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本文介绍了一种计算三维空间中凸包体积的方法。通过定义结构体来表示顶点和面，并利用向量运算来判断点的位置关系，最终实现对三维凸包体积的有效计算。

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求三维凸包，其中求体积的部分值得借鉴

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
#define MAXN 10005
const double eps = 1e-9;
int n, fc_cnt, leg[MAXN][MAXN];
double mx_ff, dis_ff;

inline int sig(double x)
{
    return (x>eps) - (x<-eps);
}
struct node
{
    double x, y, z;
    node(double a = 0.0, double b = 0.0, double c = 0.0):x(a),y(b),z(c) {}
    
    node operator - (node& a) const
    {
        return node(x-a.x, y-a.y, z-a.z);
    }
    node operator + (node& a) const
    {
        return node(x+a.x, y+a.y, z+a.z);
    }
    node operator ^ (node a) const // this * a * sin($)
    {
        return node(y*a.z-a.y*z, z*a.x-a.z*x, x*a.y-a.x*y);
    }
    double operator * (node a) const // this * a * cin($)
    {
        return x*a.x + y*a.y + z*a.z;
    }
    node cross(node a, node b)
    {
        return a - *this^b - *this;
    }
    double L()
    {
        return sqrt(x*x + y*y + z*z);
    }
    double vv6(node a, node b)
    {
        return (a^b)*(*this);
    }
} P[MAXN];
struct FACE
{
    int a, b, c;
    int isOK;
    FACE(int x = 0, int y = 0, int z = 0, int ik = 0):a(x),b(y),c(z), isOK(ik){}
    int visit(node mm)
    {
        return sig(P[a].cross(P[b], P[c])*(mm - P[a])) > 0;
    }
}facc[MAXN*8];
void deal(int o, int a, int b);
void dfs(int f, int o)
{
    facc[f].isOK = 0;
    dis_ff -= P[facc[f].a].vv6(P[facc[f].b], P[facc[f].c]);
    deal(o, facc[f].b, facc[f].a);
    deal(o, facc[f].a, facc[f].c);
    deal(o, facc[f].c, facc[f].b);
}
void deal(int o, int a, int b)
{
    int i = leg[a][b];
    if (facc[i].isOK)
    {
        if (facc[i].visit(P[o]))
            dfs(i, o);
        else
        {
            dis_ff += P[b].vv6(P[a], P[o]);
            FACE add(b, a, o, 1);
            leg[b][a] = leg[a][o] = leg[o][b] = fc_cnt;
            facc[fc_cnt++] = add;
        }
    }
}
void _update(int idx)
{
    for (int i = 0; i< fc_cnt; i++)
    {
        if (facc[i].isOK && facc[i].visit(P[idx]))
        {
            dfs(i, idx);
            break;
        }
    }
}
double getV()
{
    double v = 0;
    for (int i = 0; i<fc_cnt; i++)
    {
        if (facc[i].isOK)
        {
            v += P[facc[i].a].vv6(P[facc[i].b], P[facc[i].c]);
        }
    }
    return v/6;
}
void solve()
{
    mx_ff = dis_ff = 0;
    int kg = 1, flg = 1;
    int i, j;
    for (i = 1; i< n; i++)
    {
        if (flg == 1)
        {
            flg += sig((P[0]-P[i]).L());
            if (flg > 1)
                swap(P[1], P[i]);
        }
        else if (flg == 2)
        {
            flg += sig((P[0].cross(P[1], P[i])).L());
            if (flg > 2)
                swap(P[i], P[2]);
        }
        else if (flg == 3)
        {
            flg += sig(P[0].cross(P[1], P[2]) * (P[i]-P[0])) != 0;
            if (flg > 3)
            {
                swap(P[3], P[i]);
                fc_cnt = 0;
                for (j = 0; j< 4; j++)
                {
                    FACE add((j+1)%4, (j+2)%4, (j+3)%4, 1);
                    if (add.visit(P[j]))
                        swap(add.b, add.c);
                    leg[add.a][add.b] = leg[add.b][add.c] = leg[add.c][add.a] = fc_cnt;
                    facc[fc_cnt++] = add;
                }
                for (j = 4; j<= i; j++)
                {
                    _update(j);
                }
                kg = i+1;
                mx_ff = getV();
            }
        }
        else
        {
            dis_ff = 0;
            _update(i);
            dis_ff /= 6;
            if (sig(dis_ff - mx_ff) > 0)
            {
                mx_ff = dis_ff;
                kg = i+1;
            }
        }
    }
    printf("%d %.2lf\n", kg, mx_ff);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif
    int cs = 0;
    while (scanf("%d", &n) != EOF)
    {
        for (int i = 0; i< n; ++i)
        {
            scanf("%lf%lf%lf", &P[i].x, &P[i].y, &P[i].z);
        }
        printf("Case #%d:\n", ++cs);
        solve();
    }
    return 0;
}