Radar Installation(贪心）

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 56826		Accepted: 12814

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

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<math.h>
using namespace std;
struct island
{
    double x , y ;
    double left , right ;
}a[2000];
int n ;
int d ;
bool cmp (island a , island b)
{
    return a.x < b.x ;
}

int main ()
{
  //  freopen ("a.txt" , "r" , stdin) ;
    int cnt ;
    int ans = 1 ;
    bool flag ;

    while (~ scanf ("%d%d" , &n , &d)) {
            if (n == 0 && d == 0)
                break ;
            flag = 0 ;

        for (int i = 0 ; i < n ; i++) {
            scanf ("%lf%lf" , &a[i].x , &a[i].y) ;
            if (a[i].y > 1.0 * d || a[i].y < 0 || d <= 0) {
                flag = 1 ;
            }
            if (!flag) {
                a[i].left = (double) a[i].x - sqrt (1.0 * d * d - a[i].y * a[i].y) ;
                a[i].right = (double) a[i].x + sqrt (1.0 * d * d - a[i].y * a[i].y) ;
            }
        }
        if (flag) {
            printf ("Case %d: -1
" , ans++) ;
            continue ;
        }
        sort (a , a + n , cmp) ;
        cnt = 1 ;
        double l = a[0].left , r = a[0].right ;
        for (int i = 1 ; i < n ; i++) {
                    if (a[i].left > r) {
                            cnt++ ;
                            l = a[i].left ;
                            r = a[i].right ;
                        }
                    else {
                        l = a[i].left ;
                        r = a[i].right < r ? a[i].right : r ;
                    }
                }
        printf ("Case %d: %d
" , ans++ , cnt) ;
    }
    return 0 ;
}

别忘记每次都要跟新放radar的区间 ，orz