哈理工2015暑假训练赛 zoj 2078Phone Cell

Phone Cell Time Limit:10000MS    Memory Limit:32768KB    64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 2978

Description

Nowadays, everyone has a cellphone, or even two or three. You probably know where their name comes from. Do you. Cellphones can be moved (they are "mobile") and they use wireless connection to static stations called BTS (Base Transceiver Station). Each BTS covers an area around it and that area is called a cell.

The Czech Technical University runs an experimental private GSM network with a BTS right on top of the building you are in just now. Since the placement of base stations is very important for the network coverage, your task is to create a program that will find the optimal position for a BTS. The program will be given coordinates of "points of interest". The goal is to find a position that will cover the maximal number of these points. It is supposed that a BTS can cover all points that are no further than some given distance R. Therefore, the cell has a circular shape.

The picture above shows eight points of interest (little circles) and one of the possible optimal BTS positions (small triangle). For the given distance R, it is not possible to cover more than four points. Notice that the BTS does not need to be placed in an existing point of interest.

Input

The input consists of several scenarios. Each scenario begins with a line containing two integer numbers N and R. N is the number of points of interest, 1 <= N <= 2000. R is the maximal distance the BTS is able to cover, 0 <= R < 10000. Then there are N lines, each containing two integer numbers X_i, Y_i giving coordinates of the i-th point, |X_i|, |Y_i| < 10000. All points are distinct, i.e., no two of them will have the same coordinates.

The scenario is followed by one empty line and then the next scenario begins. The last one is followed by a line containing two zeros.

A point lying at the circle boundary (exactly in the distance R) is considered covered. To avoid floating-point inaccuracies, the input points will be selected in such a way that for any possible subset of points S that can be covered by a circle with the radius R + 0.001, there will always exist a circle with the radius R that also covers them.

Output

For each scenario, print one line containing the sentence "It is possible to cover M points.", where M is the maximal number of points of interest that may be covered by a single BTS.

Sample Input

8 2
1 2
5 3
5 4
1 4
8 2
4 5
7 5
3 3

2 100
0 100
0 -100
 
0 0

Sample Output

It is possible to cover 4 points.
It is possible to cover 2 points.

Notes

The first sample input scenario corresponds to the picture, providing that the X axis aims right and Y axis down.

本来模版中有个n^2算法的。可是超时。赛后百度了一下。还有nlogn算法的。

那么也能够当作模版来用了。

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

using namespace std ;

const double eps = 1e-8 ;
const double PI = acos(-1.0) ;
int n ;
double r ;

struct point
{
    double x,y ;
}p[2010] ;
struct node
{
    double angle ;
    int flag ;
}q[4030] ;

inline int dcmp(double d)
{
    return d < -eps ? -1 : d > eps ;
}
bool cmp(const node &a,const node &b)//角度区间排序
{
    if(dcmp(a.angle-b.angle) == 0 ) return a.flag > b.flag ;
    return a.angle < b.angle ;
}
double Sqrt(double x)
{
    return x*x ;
}
double dist(const point &a,const point &b)
{
    return sqrt(Sqrt(a.x-b.x)+Sqrt(a.y-b.y)) ;
}

int main()
{
    while(~scanf("%d %lf",&n,&r))
    {
        if(n == 0) break ;
        for(int i = 0 ; i < n ; i++)
            scanf("%lf %lf",&p[i].x,&p[i].y) ;
        int ans = 0 ;
        for(int i = 0 ; i < n ; i++)
        {
            int m = 0 ;
            for(int j = 0 ; j < n ; j++)
            {
                if(i == j) continue ;
                double d = dist(p[i],p[j]) ;
                if(d > 2*r+0.001) continue ;
                double s = atan2(p[j].y-p[i].y,p[j].x-p[i].x) ;
                if(s < 0) s += 2*PI ;//角度区间修正
                double ph = acos(d/2.0/r) ;//圆心角转区间

                q[m++].angle = s - ph + 2*PI ;q[m-1].flag = 1 ;
                q[m++].angle = s + ph + 2*PI ;q[m-1].flag = -1 ;
            }
            sort(q,q+m,cmp) ;
            int sum = 0 ;
            for(int j = 0 ; j < m ; j++)
            {
                ans = max(ans,sum += q[j].flag) ;

            }
        }
        printf("It is possible to cover %d points.\n",ans+1) ;
    }
return 0 ;
}

转载于:https://www.cnblogs.com/jhcelue/p/6941452.html