hdu 1140(三维)

War on Weather

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 494    Accepted Submission(s): 270

Problem Description

After
an unprovoked hurricane attack on the south shore, Glorious Warrior has
declared war on weather. The first salvo in this campaign will be a
coordinated pre-emptive attack on as many tropical depressions as
possible. GW reckons that the attack will neutralize the tropical
depressions before they become storms, and dissuade others from forming.

GW has at his disposal k space-to-earth killer satellites at
various locations in space. m tropical depressions are known to exist at
various locations on the earth's surface. Each satellite can attack any
number of targets on the earth provided there is line of sight between
the satellite and each target. How many different targets can be hit?

 

Input

The
input consists of several test cases. Each case begins with a line
containing integers 0 < k, m &le 100 as defined above. k lines
follow, each giving x,y,z - the location in space of a satellite at the
scheduled time of attack. m lines then follow, each giving x,y,z - the
location of a target tropical depression. Assume the earth is a sphere
centred at (0,0,0) with circumference 40,000 km. All targets will be on
the surface of the earth (within 10-9 km) and all satellites will be at
least 50 km above the surface. A line containing 0 0 follows the last
test case.

 

Output

For
each test case, output a line giving the total number of targets that
can be hit. If a particular target falls within 10-8 km of the boundary
between being within line-of-sight and not, it may be counted either
way. (That is, you need not consider rounding error so long as it does
not exceed 10-8 km.)

 

Sample Input

3 2
-10.82404031 -1594.10929753 -6239.77925152
692.58497298 -5291.64700245 4116.92402298
3006.49210582 2844.61925179 5274.03201053
2151.03635167 2255.29684503 5551.13972186
-1000.08700886 -4770.25497971 4095.48127333
3 4
0 0 6466.197723676
0 6466.197723676 0
6466.197723676 0 0
6366.197723676 0 0
6365.197723676 112.833485488 0
0 0 6366.197723676
0 -6366.197723676 0
0 0

 

Sample Output

2
3

 

刚开始被题目和AC的人吓唬了，感觉很难，，结果就一水题

题意:给出n颗卫星和m个人的三维坐标,问这m个人中有多少人可以被卫星至少一颗卫星覆盖
题解:求出卫星和地球的切线长度,在地球上与卫星的距离不超过切线长度的都能看到.

#include <iostream>
#include <cstdio>
#include <string.h>
#include <math.h>
#include <algorithm>
const double pi = atan(1.0)*;
const double R = /pi; ///地球半径
struct Point{
    double x,y,z;
}sate[],person[];

double dis(Point a,Point b){
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)+(a.z-b.z)*(a.z-b.z);
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF,n+m){
        for(int i=;i<n;i++){
            scanf("%lf%lf%lf",&sate[i].x,&sate[i].y,&sate[i].z);
        }
        for(int i=;i<m;i++){
            scanf("%lf%lf%lf",&person[i].x,&person[i].y,&person[i].z);
        }
        int ans = ;
        for(int i=;i<m;i++){
            for(int j=;j<n;j++){
                Point p = {,,}; ///球心
                int L = dis(sate[j],p)-R*R; ///切线长度
                if(sqrt(L)>=sqrt(dis(sate[j],person[i]))){
                    ans++;
                    break;
                }
            }
        }
        printf("%d\n",ans);
    }
    return ;
}