poj1328 Radar Installation 贪心

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. 

 

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

Source

Beijing 2002

# include<cstdio># include<algorithm># include<cmath>struct Section{double begin;double end;};bool mycmp(Section a,Section b){return a.begin < b.begin;}int n;//记录区间小岛的数目int d;//记录雷达的扫描半径Section section[1010];//记录每个小岛需要的雷达区间 bool init(){bool legalData = true;int x,y;for(int i=0;i<n;i++){scanf("%d%d",&x,&y);section[i].begin = x-sqrt(d*d -y*y);section[i].end =x+sqrt(d*d -y*y);if(y>d)legalData =false;}std::sort(section,section+n,mycmp);return legalData;}int main(){//freopen("in","r",stdin);//freopen("out","w",stdout);int cas =1;while(scanf("%d %d",&n,&d)!=EOF&&(n||d)){if(! init()){printf("Case %d: -1\n",cas);cas++;continue;}if(n==1){printf("Case %d: 1\n",cas);cas++;continue;}double pre =section[0].end;Section *now;int num =1;for(int i=1;i<n;i++){now =&(section[i]) ;if(now->begin > pre)   //两个区间不重叠 {                    num++;pre =now->end;}else if(now->end <pre){pre = now->end;}}printf("Case %d: %d\n",cas,num);cas++;}return 0;}