Radar Installation（贪心 区间覆盖）

Radar Installation
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u
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Status
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 2
1 2
-3 1
2 1

1 2
0 2

0 0
Sample Output
Case 1: 2
Case 2: 1

题意：多组数据输入，每组数据第一行给出N，R接着有N组小岛坐标，雷达半径为R。X轴以上是海洋，X轴以下是陆地，雷达只能部署到X轴上，输出覆盖海洋上N个小岛最少需要多少雷达，如果不能完全覆盖，输出-1；

思路：假设雷达在小道上，那么雷达覆盖范围和X轴就有相交区间，记录下这个区间的左右极限，接着就是区间覆盖问题，有重叠区间的部分只要一个雷达就够了，如果发现有雷达覆盖范围不和X轴相交，那就输出-1吧

代码

#include<stdio.h>#include<iostream>#include<algorithm>#include<string.h>#include<math.h>#include<queue>using namespace std;const int maxn=1005;//区间覆盖问题struct node{    double left;    double right;} point[maxn]; //记录小岛雷达站在陆地覆盖的区间bool cmp(node x,node y){    if(x.right==y.right)        return x.left<y.left;    return x.right<y.right;}int main(){    int N;//岛屿数量    double D;//雷达半径    int casen=1;    while(~scanf("%d%lf",&N,&D)&&N&&D)    {        bool flag=0;        for(int i=0; i<N; i++)        {            double x,y;            scanf("%lf%lf",&x,&y);            if(y>D)                flag=1;            point[i].left=x-sqrt(D*D-y*y);            point[i].right=x+sqrt(D*D-y*y);        }        if(flag==1)        {            printf("Case %d: -1\n",casen++);            continue;        }        sort(point,point+N,cmp);        int result=1;        double right=point[0].right;        for(int i=1;i<N;i++)        {            if(point[i].left>right)            {                result++;                right=point[i].right;            }        }        printf("Case %d: %d\n",casen++,result);    }    return 0;}