POJ1328——贪心——Radar Installation

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

Source

Beijing 2002

大意：河边有很多岛，要在沿岸建雷达，将点化成在沿岸的坐标，进行贪心，注意当排序完左之后，要考虑如果有点的右在ax的右，那么ax就变成该店，因为这个，一直没A~~~

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int maxn = 1000;
struct X{
    double left;
    double right;
}dp[maxn];
int x[maxn],y[maxn];
bool cmp(X i,X j){
return i.left<j.left;}
int main(){
    int n,d,i, flag1 = 1;
    while(~scanf("%d%d",&n,&d)&&n&&d){
            int flag2 = 0;
            for( i = 1; i <= n ; i++){
              scanf("%d%d",&x[i],&y[i]);
              if(y[i] > d||d<0||y[i]<0)
              flag2 = 1;
             dp[i].left = x[i] - sqrt((double)(d*d-y[i]*y[i]));
             dp[i].right = x[i] + sqrt((double)(d*d-y[i]*y[i]));
            }
            if(flag2 == 1) {
                    printf("Case %d: -1\n",flag1);
                    flag1++;
            continue;}
            sort(dp+1,dp+n+1,cmp);
    int ax = 1,flag = 1;
        for(i = ax + 1; i <= n ; i++){
            if(dp[i].right < dp[ax].right){
                    ax = i;
            }
             else  if(dp[i].left > dp[ax].right){
                ax = i;
                flag++;
              }
          }
        printf("Case %d: %d\n",flag1,flag);
        flag1++;
     }
    return 0;
}