POJ Problem Radar Installation 【贪心】

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 75347		Accepted: 16863

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

海上的每一个岛到陆地上最远距离d有两个点，当雷达在这两个点所确定的区域时，这个岛就在雷达所在的范围内。所以就将这个问题化为贪心的区间覆盖问题。将所有岛的区间左端点进行升序排列，每次使右端点所能包含的线段数目最大（答案为统计左右不重合的线段的数目）。

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAX_N 1010
#define MAX(a, b)   ((a > b)? a: b)
#define MIN(a, b)   ((a < b)? a: b)
using namespace std;
const int INF = 1e8;

struct node{
    double l, r;
}pon[MAX_N];

bool cmp(node x, node y) {
    return x.l < y.l;
}
int main() {
    int n, d;
    double x, y;
    int cnt = 0;
    while (scanf("%d%d", &n, &d), n||d) {
        int ans = 0;
        bool flag = false;
        for (int i = 0; i < n; i++) {
            scanf("%lf%lf", &x, &y);
            if (y > d) {
                flag = true;
            }
            pon[i].l = x - sqrt(d*d - y*y);
            pon[i].r = x + sqrt(d*d - y*y);
        }
        if (flag) {
            printf("Case %d: -1
", ++cnt);
            continue;
        }
        sort(pon, pon + n, cmp);
        double t = pon[0].r;
        for (int i = 1; i < n; i++) {
            if(pon[i].l > t) {
				t = pon[i].r;
				ans++;
			}
			//如果当前的右端点大于pon[i].r，则当前的右端点应等于pon[i].r
            else if(pon[i].r < t) {
                t = pon[i].r;
            }
        }
        printf("Case %d: %d
", ++cnt, ans + 1);
    }
    return 0;
}