nyoj Radar

描述

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.

 

输入

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

输出

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.

样例输入

3 21 2-3 12 11 20 20 0

样例输出

Case 1: 2Case 2: 1

贪心算法，先将所有的点转化成区间，对区间左点排序，比较。。。代码：#include<stdio.h>#include<math.h>#include<algorithm>using namespace std;struct readar{double left;double right;}s[1500];bool cmp(readar x,readar y){if(x.right<y.right)   return true;return false;} int main(){int n;double a,r,b;int i,j,k;double num,t;int cas=1;while(~scanf("%d%lf",&n,&r),n!=0&&r!=0){int flag=0;for(i=0;i<n;i++){scanf("%lf%lf",&a,&b);if(r<b){flag=1;continue;}num=sqrt(r*r-b*b);s[i].left=a-num;s[i].right=a+num;}sort(s,s+n,cmp);int ans=1;t=s[0].right;for(i=1;i<n;i++){if(s[i].left>t){ans++;t=s[i].right;}} if(flag)printf("Case %d: -1\n",cas++);elseprintf("Case %d: %d\n",cas++,ans);} return 0;}