大意:中文题,很好理解,搞清楚各种变量就行。
思路:我知道的好像有两种解法,一种是求土匪的头心与子弹射出的直线求点到直线距离,在判断一下方向对不对;另一种是求出子弹射出点与土匪头心连线,求出子弹的射出的直线,求两直线的夹角, 求出子弹射出点与土匪头心连线,求出求出子弹射出点与土匪头的切线,求两直线的夹角,比较这两个夹角的大小判断是不是会打到。
这里我用第一种方法过的,就贴第一种的吧。
1 struct point 2 { 3 double x, y, z; 4 } A, B, C; 5 6 ///计算cross product U x V 7 point xmult(point u,point v){ 8 point ret; 9 ret.x=u.y*v.z-v.y*u.z;10 ret.y=u.z*v.x-u.x*v.z;11 ret.z=u.x*v.y-u.y*v.x;12 return ret;13 }14 15 ///两点距离,单参数取向量大小16 double Distance(point p1,point p2){17 return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)+(p1.z-p2.z)*(p1.z-p2.z));18 }19 20 ///矢量差 U - V21 point subt(point u,point v){22 point ret;23 ret.x=u.x-v.x;24 ret.y=u.y-v.y;25 ret.z=u.z-v.z;26 return ret;27 }28 29 ///向量大小30 double vlen(point p){31 return sqrt(p.x*p.x+p.y*p.y+p.z*p.z);32 }33 34 double ptoline(point p,point l1,point l2){35 return vlen(xmult(subt(p,l1),subt(l2,l1)))/Distance(l1,l2);36 }37 38 int n;39 double h1,r1;40 double h2,r2,x3,y3,z3;41 42 void Solve()43 {44 scanf("%d", &n);45 while(n--)46 {47 scanf("%lf%lf%lf%lf%lf", &h1, &r1, &A.x, &A.y, &A.z);48 scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &h2, &r2, &B.x, &B.y, &B.z, &C.x, &C.y, &C.z);49 50 A.z = A.z+h1-r1;51 B.z = B.z+h2*0.9-r2;52 53 double x = A.x-B.x;54 double y = A.y-B.y;55 double z = A.z-B.z;56 //printf("%lf %lf %lf\n", x, y, z);57 58 point D;59 60 D.x = C.x+B.x;61 D.y = C.y+B.y;62 D.z = C.z+B.z;63 64 double d = ptoline(A, B, D);65 //printf("%lf\n", d);66 67 if(d <= r1 && (x*C.x+y*C.y+z*C.z > 0))68 {69 printf("YES\n");70 }71 else72 {73 printf("NO\n");74 }75 }76 }