poj Radar Installation 【贪心 区间】

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 75358 Accepted: 16865

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

Source

Beijing 2002

嗯，每个岛屿的雷达有个区间范围，区间覆盖，[l , r], 右端点从小到大排序，并更新右端点

坐标是浮点型

rightn=-1000000000

#include<cstdio>#include<cmath> #include<algorithm> #define INF 1000000000using namespace std;struct Node{double left,right;}node[1010];bool cmp(Node x,Node y){return x.right<y.right;}int main(){int n,k=0;double d,x,y;while(scanf("%d%lf",&n,&d)&&(n||d)){int flag=1;for(int i=0;i<n;i++){scanf("%lf%lf",&x,&y);if(!flag)continue;if(y>d){flag=0;continue;}node[i].left=x-sqrt(d*d-y*y);node[i].right=x+sqrt(d*d-y*y);}printf("Case %d: ",++k);//先输出Case if(!flag){printf("-1\n");continue;}sort(node,node+n,cmp);double rightn=-INF;//横坐标是浮点型数 int ans=0;for(int i=0;i<n;i++){if(rightn<node[i].left){ans++;rightn=node[i].right;}}printf("%d\n",ans); }  return 0;}