E - Shooting Game FZU - 2144
Fat brother and Maze are playing a kind of special (hentai) game in the playground. (Maybe it’s the OOXX game which decrypted in the last problem, who knows.) But as they don’t like using repellent while playing this kind of special (hentai) game, they really suffer a lot from the mosquito. So they decide to use antiaircraft gun to shoot the mosquito. You can assume that the playground is a kind of three-dimensional space and there are N mosquitoes in the playground. Each of them is a kind of point in the space which is doing the uniform linear motion. (匀速直线运动) Fat brother is standing at (0, 0, 0) and once he shoot, the mosquito who’s distance from Fat brother is no large than R will be shot down. You can assume that the area which Fat brother shoot is a kind of a sphere with radio R and the mosquito inside this sphere will be shot down. As Fat brother hate these mosquito very much, he wants to shoot as much mosquito as he can. But as we all know, it’s tired for a man to shoot even if he is really enjoying this. So in addition to that, Fat brother wants to shoot as less time as he can.
You can (have to) assume that Fat brother is strong enough and he don’t need to rest after shooting which means that can shoot at ANY TIME.
Input
The first line of the date is an integer T, which is the number of the text cases.
Then T cases follow, each case starts with two integers N and R which describe above.
Then N lines follow, the ith line contains six integers ax, ay, az, dx, dy, dz. It means that at time 0, the ith mosquito is at (ax, ay, az) and it’s moving direction is (dx, dy, dz) which means that after time t this mosquito will be at (ax+dx*t, ay+dy*t, ax+dz*t). You can assume that dx*dx + dy*dy+ dz*dz > 0.
1 <= T <= 50, 1 <= N <= 100000, 1 <= R <= 1000000
-1000000 <= ax, ay, az <= 1000000
-100 <= dx, dy, dz <= 100
The range of each coordinate is [-10086, 10086]
For each case, output the case number first, then output two numbers A and B.
A is the number of mosquito Fat brother can shoot down.
B is the number of times Fat brother need to shoot.
6 2 1 2 0 0 -1 0 0 -2 0 0 1 0 0 2 1 4 0 0 -1 0 0 -2 0 0 1 0 0 2 1 4 0 0 -1 0 0 1 0 0 1 0 0 2 1 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 1 1 0 0 0 1 0 0 3 1 -1 0 0 1 0 0 -2 0 0 1 0 0 4 0 0 -1 0 0Sample Output
Case 1: 2 1 Case 2: 2 1 Case 3: 2 2 Case 4: 0 0 Case 5: 1 1 Case 6: 3 2
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
typedef long long ll;
using namespace std;const double eps=1e-9;
const int maxn=1e5+10;
struct node{double x,y,z;double vx,vy,vz;node(){ }node(double _x,double _y,double _z,double _vx,double _vy,double _vz){x=_x,y=_y,z=_z,vx=_vx,vy=_vy,vz=_vz;}
}p[maxn];
struct SEG{double st,ed;bool operator<(const SEG& b)const{if(fabs(st-b.st)<=eps)return ed>b.ed; return st<b.st;}
}arr[maxn];
/*
注意无解的情况
*/
int check(int i,double& st,double& ed,double R){double A=(p[i].vx*p[i].vx+p[i].vy*p[i].vy+p[i].vz*p[i].vz),B=2.0*(p[i].x*p[i].vx+p[i].y*p[i].vy+p[i].z*p[i].vz),C=(p[i].x*p[i].x+p[i].y*p[i].y+p[i].z*p[i].z)-R*R;if(B*B-4.0*A*C<0) return 0;//忘了没解的情况 double ans2=(-B+sqrt(B*B-4.0*A*C))/(2.0*A);if(ans2<0) return 0;double ans1=max(0.0,(-B-sqrt(B*B-4.0*A*C))/(2.0*A));st=ans1,ed=ans2;//printf("A:%f B:%f C:%f\n",A,B,C);return 1;
}/*
??????卡时间，直接按照doube输入超时，得用int再转化。
*/
int main(){int T,kase=1;scanf("%d",&T);while(T--){int n;double r;scanf("%d %lf",&n,&r);for(int i=1;i<=n;i++){int x,y,z,dx,dy,dz;scanf("%d %d %d %d %d %d",&x,&y,&z,&dx,&dy,&dz);p[i]=node(x*1.0,y*1.0,z*1.0,dx*1.0,dy*1.0,dz*1.0);}int cnt=0;for(int i=1;i<=n;i++){double st,ed;if(check(i,st,ed,r)){arr[cnt].st=st,arr[cnt++].ed=ed;}} if(cnt==0){printf("Case %d: %d %d\n",kase++,0,0);continue;}sort(arr,arr+cnt);int ans=1;double max_line=arr[0].ed;for(int i=1;i<cnt;i++){if(arr[i].st>max_line){ans++;max_line=arr[i].ed;}max_line=min(max_line,arr[i].ed);}printf("Case %d: %d %d\n",kase++,cnt,ans);} return 0;
}#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
typedef long long ll;
using namespace std;const double eps=1e-9;
const int maxn=1e5+10;
struct node{double x,y,z;double vx,vy,vz;node(){ }node(double _x,double _y,double _z,double _vx,double _vy,double _vz){x=_x,y=_y,z=_z,vx=_vx,vy=_vy,vz=_vz;}
}p[maxn];
struct SEG{double st,ed;bool operator<(const SEG& b)const{return ed<b.ed;}
}arr[maxn]; int check(int i,double& st,double& ed,double R){double A=(p[i].vx*p[i].vx+p[i].vy*p[i].vy+p[i].vz*p[i].vz),B=2.0*(p[i].x*p[i].vx+p[i].y*p[i].vy+p[i].z*p[i].vz),C=(p[i].x*p[i].x+p[i].y*p[i].y+p[i].z*p[i].z)-R*R;if(B*B-4.0*A*C<0) return 0;//忘了没解的情况 double ans2=(-B+sqrt(B*B-4.0*A*C))/(2.0*A);if(ans2<0) return 0;double ans1=max(0.0,(-B-sqrt(B*B-4.0*A*C))/(2.0*A));st=ans1,ed=ans2;//printf("A:%f B:%f C:%f\n",A,B,C);return 1;
}
/*
4
2 1
1 1 1 1 1 1
-1 -1 -1 -1 -1 -1
1 1
2 0 0 0 1 0
*//*
??????卡时间，直接按照doube输入超时，得用int再转化。
*/
int main(){int T,kase=1;scanf("%d",&T);while(T--){int n;double r;scanf("%d %lf",&n,&r);for(int i=1;i<=n;i++){int x,y,z,dx,dy,dz;scanf("%d %d %d %d %d %d",&x,&y,&z,&dx,&dy,&dz);p[i]=node(x*1.0,y*1.0,z*1.0,dx*1.0,dy*1.0,dz*1.0);}int cnt=0;for(int i=1;i<=n;i++){double st,ed;if(check(i,st,ed,r)){arr[cnt].st=st,arr[cnt++].ed=ed;// printf("st:%f ed:%f\n",st,ed);}} if(cnt==0){printf("Case %d: %d %d\n",kase++,0,0);continue;}sort(arr,arr+cnt);int ans=1;double max_line=arr[0].ed;for(int i=1;i<cnt;i++){if(arr[i].st>max_line){ans++;max_line=arr[i].ed;}}printf("Case %d: %d %d\n",kase++,cnt,ans);} return 0;
}