Rader Installation(POJ_1328)

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

代码

一开始贪心的方法想错了，后来的方法写的时候也是错误百出，目测是多年没刷题的后果233。我的思路是，如果两个点所形成的覆盖区域截取的x轴上的线段有重叠的话，那么这两个点一定能被同一个x轴上的点覆盖。就先计算出各点的覆盖圆在x轴上截取线段的左右坐标，按照左坐标从小到大排一下序，然后贪心就行了。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;

struct node
{
    double L, R;
    bool operator <(const node b)const
    {
        return L < b.L;
    }
    struct node &operator =(const node b)
    {
        L = b.L;
        R = b.R;
        return *this;
    }
}a[1100];

int main()
{
    int n, d;
    double x, y;
    int cnt = 0;
    while(scanf("%d %d", &n, &d) && (n || d))
    {
        bool flag = 0;
        cnt++;
        for(int i = 0; i < n; i++)
        {
            scanf("%lf %lf", &x, &y);
            if(y > d || y < -d)
                flag = 1;
            a[i].L = x - sqrt(d * d - y * y);
            a[i].R = x + sqrt(d * d - y * y);
        }
        if(flag)
        {
            printf("Case %d: -1\n",cnt);
            continue;
        }
        sort(a, a + n);
        int ans = 1;
        struct node k;
        k = a[0];
        for(int i = 1; i < n; i++)
        {
            if(a[i].R < k.R)
                k = a[i];
            else if(a[i].L > k.R)
            {
                ans++;
                k = a[i];
            }
        }
        printf("Case %d: %d\n", cnt, ans);
    }
    return 0;
}