poj 1328 Radar Installation

Description

Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d.

We use Cartesian coordinate system, defining the coasting is the x-axis. The sea side is above x-axis, and the land side below. Given the position of each island in the sea, and given the distance of the coverage of the radar installation, your task is to write a program to find the minimal number of radar installations to cover all the islands. Note that the position of an island is represented by its x-y coordinates.

Figure A Sample Input of Radar Installations

Input

The input consists of several test cases. The first line of each case contains two integers n (1<=n<=1000) and d, where n is the number of islands in the sea and d is the distance of coverage of the radar installation. This is followed by n lines each containing two integers representing the coordinate of the position of each island. Then a blank line follows to separate the cases.

The input is terminated by a line containing pair of zeros

Output

For each test case output one line consisting of the test case number followed by the minimal number of radar installations needed. "-1" installation means no solution for that case.

思路：

以小岛为圆心，以d为半径，找出和x轴的两个交点l和r。

统计所有小岛的l和r，并按照r排序。

(活动选择问题)

首先选择第一个r，后去掉所有l < 该r的小岛(活动)

#include<iostream>
#include<math.h>
#include<memory.h>
using namespace std;
double sourceXY[1000][2];
double X[1010][2];
double maxRadius;
int partition2(double sourceXY[][2],int p,int r){
    double baseX = sourceXY[p][0];
    double baseY = sourceXY[p][1];
    while(p < r){
        while(p < r && ((sourceXY[r][0] > baseX) || (sourceXY[r][0] == baseX && sourceXY[r][1] >= baseY)) )
            r--;
        if(p < r){
            sourceXY[p][0] = sourceXY[r][0];
            sourceXY[p][1] = sourceXY[r][1];
            p++;
        }
        while(p < r &&  ( (sourceXY[p][0] < baseX) || (sourceXY[p][0] == baseX && sourceXY[p][1] <= baseY ) ) )
            p++;
        if(p < r){
            sourceXY[r][0] = sourceXY[p][0];
            sourceXY[r][1] = sourceXY[p][1];
            r--;
        }
    }
    sourceXY[p][0] = baseX;
    sourceXY[p][1] = baseY;
    return p;
}
void sortByX(double sourceXY[][2],int p,int r){
    //快排把
    if(p < r){
        int q = partition2(sourceXY,p,r);
        sortByX(sourceXY,p,q-1);
        sortByX(sourceXY,q+1,r);
    }
}
int main(){
    int cases;
    int flag = 1;
    int numOfCases = 1;
    while(cin>>cases>>maxRadius && cases != 0){
        flag = 1;
        for(int i = 0; i < cases;i++){
            cin>>sourceXY[i][0]>>sourceXY[i][1];
            if(sourceXY[i][1] > maxRadius){
                flag = 0;
            }
        }
        if(maxRadius <= 0 ||flag == 0 ||cases <= 0){
            cout<<"Case "<<numOfCases<<": -1"<<endl;
            numOfCases++;
        }else{


            for(int i = 0; i < cases; i++){
                double temp = sqrt(maxRadius*maxRadius*1.0 - sourceXY[i][1]*sourceXY[i][1]*1.0);
                X[i][1] = sourceXY[i][0]-temp;
                X[i][0] = temp +sourceXY[i][0];
            }
            sortByX(X,0,cases-1);
            int k = 0,t=1;
            int results = 1;
            while(t < cases){

                while(t < cases && X[t][1] <= X[k][0])
                    t++;
                if(t == cases)
                    break;
                k = t;
                results++;
                t++;
            }
            cout<<"Case "<<numOfCases<<": "<<results<<endl;
            numOfCases++;
        }
    }
}