不同屬性不繫結，可以使用貪心法

如果繫結，則不能貪心法解決

Radar Installation

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

思路：轉換成區間選點問題，用貪心演算法即可解決問題！！！即按照右端點進行對區間從小到大排序，之後詳細看下面的程式碼中的貪心演算法部分！！！

AC程式碼：

被坑了，特浪費一個上午時間！！！！！！！！！！！！！！！！！！！！！！！！！！！！

Then a blank line follows to separate the cases. 被這句話坑慘了，本題目在輸入時候要求多輸入一個空行，如果多寫一個printf("\n");則無法通過，謹記謹記！！！

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#include<cstdio> #include<cmath> #include<algorithm> using namespace std; const int maxn=1005; typedef struct { double x; double y; }point; typedef struct { double x1; double x2; }region; point p[maxn]; region r[maxn]; bool cmp(region r1,region r2) { return r1.x2<r2.x2; } int main() { int n; double d; int num=0; while(scanf("%d%lf",&n,&d)!=EOF) { if(n==0 && d==0)break; bool flag=1; for(int i=0;i<n;i++) { scanf("%lf%lf",&p[i].x,&p[i].y); double t=sqrt(d*d-p[i].y*p[i].y); r[i].x1=p[i].x-t; r[i].x2=p[i].x+t; if(p[i].y>d)flag=0; } printf("Case %d: ",++num); if(!flag){printf("-1\n");continue;} sort(r,r+n,cmp); //貪心演算法如下 double pos=r[0].x2; int cnt=1; for(int i=1;i<n;i++) { if(r[i].x1>pos) { cnt++; pos=r[i].x2; } } printf("%d\n",cnt); } return 0; }