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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 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;}