（POJ

（POJ - 1328）Radar Installation

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 90183 Accepted: 20266

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.

Sample Input

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2
Case 2: 1

题目大意：有n个小岛可视为点，海岸视为x轴，海岸上装雷达的可管辖半径是d，问最少需要多少雷达，可是所有小岛都处于管辖范围。

思路：对于每一个小岛，处理出能管辖到他的雷达得范围（即雷达可装的最左边距离和最右边距离）的，按处理出来的左边距离排序，如果有小岛能管辖他的的雷达能装的最左边位置>之前那个，那么就需要再装一个雷达，所有雷达竟可能往右边装，这样才能保证尽量少的雷达管到多有小岛。

#include<cstdio>#include<algorithm>#include<cmath>using namespace std;const int maxn=1005;struct point{    double x,y;}island[maxn];struct node{    double left,right;}radar[maxn]; bool cmp(node a,node b){    return a.left<b.left;}int main(){    int n,d,kase=1;    while(~scanf("%d%d",&n,&d)&&n+d)    {        bool flag=false;        for(int i=0;i<n;i++)        {            scanf("%lf%lf",&island[i].x,&island[i].y);            if(island[i].y>(double)d) flag=true;         }        if(flag)         {            printf("Case %d: -1\n",kase++);            continue;        }        for(int i=0;i<n;i++)        {            double dx=sqrt(d*d*1.0-island[i].y*island[i].y);            radar[i].left=island[i].x-dx;            radar[i].right=island[i].x+dx;        }        sort(radar,radar+n,cmp);        int ans=1;        double tmp=radar[0].right;        for(int i=1;i<n;i++)        {            if(radar[i].left>tmp)            {                ans++;                tmp=radar[i].right;            }            else if(radar[i].right<tmp) tmp=radar[i].right;        }        printf("Case %d: %d\n",kase++,ans);    }    return 0;}