POJ 1328：Radar Installation 经典贪心

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 67532 Accepted: 15147

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

题意是在平面上有一些卫星，要在X轴上建立一些雷达，雷达有相同的辐射半径，问想要把所有卫星都辐射到需要的最少的雷达数量。如果不可能都辐射到的话，输出-1。

2012年做的题，很经典的贪心。实质上这个问题贪的是区间，要找到最大数量重合的区间，然后再找下一个最大数量重合的区间，这时结果就+1。

代码：

#include <iostream>#include <cmath>#include <algorithm>using namespace std; struct ss {  double x,y;  }a[1005]; int cmp(ss x,ss y)  {  if(x.x==y.x) return x.y<y.y; else return x.x<y.x; }int main(){int island,count,result=0,flag=0,R,island_x[1005],island_y[1005];double radar;cin>>island>>R;while(island!=0||R!=0){for(count=1;count<=island;count++){cin>>island_x[count]>>island_y[count];if(island_y[count]>R||island_y[count]<0||R<=0){result=-1;}else{a[count].x=island_x[count]-sqrt(1.0*(R*R-island_y[count]*island_y[count]));a[count].y=island_x[count]+sqrt(1.0*(R*R-island_y[count]*island_y[count]));}}if(result!=-1){    sort(a+1,a+island+1,cmp);radar=a[1].y;result=1;for(count=2;count<=island;count++){if(a[count].x>radar)//当前区间不能满足重合的要求了，要增加一个区间了{result++;radar=a[count].y;}else if(a[count].y<radar){radar=a[count].y;//缩小重合区间}}cout<<"Case "<<++flag<<": "<<result<<endl;result=0;}else{cout<<"Case "<<++flag<<": "<<result<<endl;result=0;}cin>>island>>R;}return 0;}