Radar Installation（poj1328）（贪心）

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 54756 Accepted: 12337

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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1

/*思路：该题题意是为了求出能够覆盖所有岛屿的最小雷达数目，每个小岛对应x轴上的一个区间，在这个区间内的任何一个点放置雷达，则可以覆盖该小岛，区间范围的计算用[x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)];问题即转化为已知一定数量的区间，求最少区间可以包含所有的点，使得每个区间内斗至少存在一个点。每次迭代对于第一个区间， 选择最右边一个点， 因为它可以让较多区间得到满足，选择该点之后，将得到满足的区间删掉， 进行下一步迭代， 直到所有的点均包含则结束。不过还要考虑d<0和y>d的情况。 左端点：表示雷达能够覆盖岛左临界的雷达， 右端点：表示雷达能够覆盖岛右临界的雷达。 */#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>using namespace std;struct st{double left,right;}data[1010];int cmp(st a,st b){return a.left<b.left;}int main(){int i,t,kase=1,sum,n,d;double x,y,ans;while(scanf("%d%d",&n,&d)&&(n||d)){t=0;for(i=0;i<n;i++){    scanf("%lf %lf",&x,&y);    if(y>d||d<0)    {    t=1;    }    data[i].left=x-sqrt(d*d-y*y);    data[i].right=x+sqrt(d*d-y*y);}sort(data,data+n,cmp);if(t==1){    printf("Case %d: -1\n",kase);    kase++;}else{ans=data[0].right;sum=1;for(i=1;i<n;i++){if(data[i].right<=ans)//如果右端点在前一雷达覆盖范围，      ans=data[i].right;//则雷达要放在放在右端点。 else if(data[i].left>ans)//如果左端点不在前一雷达覆盖范围内， {                         //则需要增加新雷达，并将雷达放在右端点。     sum++;    ans=data[i].right;}}printf("Case %d: %d\n",kase,sum);kase++;}}return 0;}