HDU 4617 Weapon （计算几何：异面直线距离）

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4617

题意：给出空间内的n个圆的圆心及圆上两点，即n个无限延长的光柱的截面，问这些圆柱是否会交在一起，如果不交输出最短距离，否则输出"Lucky"

思路：对于每个圆，找出其法向量直线，这样每两条法向量直线的间距减去两圆的半径和即是距离，判断该距离是否<=0即可

#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <iostream>#include <algorithm>#include <utility>#include <functional>#include <vector>#include <queue>#include <string>#include <set>#include <map>#pragma comment (linker, "/STACK:1024000000,1024000000")#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1#define eps 1e-8#define sign(x) ((x) > eps ? 1 : ((x) < -eps ? (-1) : (0)))using namespace std;typedef long long ll;int n;struct point{    double x, y, z;    point() {}    point(double _x, double _y, double _z)    {        x = _x;        y = _y;        z = _z;    }    void input()    {        scanf("%lf%lf%lf", &x, &y, &z);    }    point operator + (const point &b) const    {        return point(x + b.x, y + b.y, z + b.z);    }};struct line{    point a, b;};struct circle{    point o, p1, p2;    void input()    {        o.input();        p1.input();        p2.input();    }} c[100];point xmult(point u, point v){    point ret;    ret.x = u.y * v.z - v.y * u.z;    ret.y = u.z * v.x - u.x * v.z;    ret.z = u.x * v.y - u.y * v.x;    return ret;}double dis(point u, point v){    return sqrt((u.x - v.x) * (u.x - v.x) + (u.y - v.y) * (u.y - v.y) + (u.z - v.z) * (u.z - v.z));}double dmult(point u, point v){    return u.x * v.x + u.y * v.y + u.z * v.z;}point sub(point u, point v){    point ret;    ret.x = u.x - v.x;    ret.y = u.y - v.y;    ret.z = u.z - v.z;    return ret;}double vlen(point p){    return sqrt(p.x * p.x + p.y * p.y + p.z * p.z);}point pvec(circle s){    return xmult(sub(s.o, s.p1), sub(s.p1, s.p2));}double linedis(line u, line v){    point n = xmult(sub(u.a, u.b), sub(v.a, v.b));    return fabs(dmult(sub(u.a, v.a), n)) / vlen(n);}void solve(){    double ans = 1e9;    for(int i = 0; i < n; i++)    {        for(int j = i + 1; j < n; j++)        {            circle c1 = c[i], c2 = c[j];            double r1 = dis(c[i].o, c[i].p1);            double r2 = dis(c[j].o, c[j].p1);            point k1 = pvec(c1);            point k2 = pvec(c2);            line l1, l2;            l1.a = c1.o, l1.b = k1 + c1.o;            l2.a = c2.o, l2.b = k2 + c2.o;            double d = linedis(l1, l2);            if(sign(d - r1 - r2) <= 0)            {                puts("Lucky");                return ;            }            ans = min(ans, d - r1 - r2);        }    }    printf("%.2f\n", ans);}int main(){    int t;    scanf("%d", &t);    while(t--)    {        scanf("%d", &n);        for(int i = 0; i < n; i++)            c[i].input();        solve();    }    return 0;}

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