FZU 2144 Shooting Game （贪心区域划分）

Problem 2144 Shooting Game

Accept: 370 Submit: 1902

Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem Description

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]

Output

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.

Sample Input

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 0

Sample 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

可以根据出去的时间对蚊子进行排序，然后统计后面有多少蚊子进去的时间是小于这个蚊子的出去时间，这些蚊子可以同时杀死。

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <algorithm>
#include <math.h>

using namespace std;
#define MAX 100000
int t;
int n,r;
struct Node
{
    double leftime;
    double getime;
}a[MAX+5];
int cmp(Node a,Node b)
{
    if(a.leftime==b.leftime)
        return a.getime<b.getime;
    return a.leftime<b.leftime;
}
int ans;
int res;
int tag[MAX+5];
int  main()
{
    int x1,y1,z1,x2,y2,z2;
    scanf("%d",&t);
    int cas=0;
    while(t--)
    {
        ans=0;
        res=0;
        scanf("%d%d",&n,&r);
        int cnt=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);
            double a1=(x2*1.0)*(x2*1.0)+(y2*1.0)*(y2*1.0)+(z2*1.0)*(z2*1.0);
            double b=2*(x1*1.0)*(x2*1.0)+2*(y1*1.0)*(y2*1.0)+2*(z1*1.0)*(z2*1.0);
            double c=(x1*1.0)*(x1*1.0)+(y1*1.0)*(y1*1.0)+(z1*1.0)*(z1*1.0)-(r*1.0)*(r*1.0);
            double d=b*b-4*a1*c;
            if(d<0) continue;
            double xx=(-b+sqrt(d))/(2*a1);
            double yy=(-b-sqrt(d))/(2*a1);
            if(xx<0&&yy<0) continue;
            a[cnt].getime=min(xx,yy);
            if(a[cnt].getime<0)  a[cnt].getime=0;;
            a[cnt++].leftime=max(xx,yy);
        }
        sort(a,a+cnt,cmp);
        ans=cnt;
        res=0;
        int k;
        for(int i=0;i<cnt;i=k)
        {
            k=i+1;
            for(int j=i+1;k<cnt;j++)
            {
                if(a[j].getime<=a[i].leftime)
                    k=j+1;
                else
                    break;
            }
            res++;

        }
        printf("Case %d: %d %d\n",++cas,ans,res);
    }
    return 0;
}