题解|#Radar Installation#(区间贪心)

Radar Installation（http://poj.org/problem?id=1328）

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.

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;

struct range{//雷达安装区间
  double left;
  double right;
};

bool compare(range x,range y){//比较函数
  return x.left<y.left;
}

range radar[1010];

int main(){
  int n,d;
  int casenum=1;
  while(scanf("%d%d",&n,&d)!=EOF){
    if(n==0&&d==0)break;
    bool flag=true;
    for(int i=0;i<n;i++){
      int x,y;
      scanf("%d%d",&x,&y);
      if(y>d)flag=false;
      else{
        radar[i].left=x-sqrt(d*d-1.0*y*y);
        radar[i].right=x+sqrt(d*d-1.0*y*y);
      }
    }
    if(!flag){
      printf("Case %d: -1\n",casenum++);
    }
    else{
      sort(radar,radar+n,compare);//左端点升序
      double cur=radar[0].right;
      int sum=1;
      for(int i=1;i<n;i++){
        if(radar[i].left<=cur){
          cur=min(cur,radar[i].right);//取重合区间的右端点
        }
        else{
          cur=radar[i].right;//再装一个雷达
          sum++;
        }
      }
      printf("Case %d: %d\n",casenum++,sum);
    }
  }
    
}