poj1328 Radar Installation（贪心）

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 46630		Accepted: 10405

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

 

题意：

在海岸线上安放最少数目的雷达使得所有小岛都能在其辐射区内

思路：

先将各个小岛可被辐射的雷达安放范围依次求出，即转化为区间选点问题（将其右端点升序排列）。注意：本题一定要考虑无解的情况！

 

#include
#include
#include
using namespace std;
struct ss
{
    int x,y;
    double left,right;
} s[1004];
int cmp(ss a,ss b)
{
    return a.rightd||d<0)
            {
                ok=0;
                break;
            }
            double temp=sqrt((double)(d*d-s[i].y*s[i].y));
            s[i].left=s[i].x-temp;
            s[i].right=s[i].x+temp;
        }
        if(!ok)
        {
            printf("Case %d: -1\n",cas++);
            continue;
        }
        sort(s,s+n,cmp);
        ans=1;
        flag=s[0].right;
        for(i=1; iflag)
            {
                flag=s[i].right;
                ans++;
            }
        }
        printf("Case %d: %d\n",cas++,ans);
    }
    return 0;
}