UVALive - 2519 Radar Installation 解题心得

原题：

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.

 

Input

The input consists of several test cases. The first line of each case contains two integers n (1n1000) 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 nlines 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:

给出一个坐标轴, y的正半轴是海, 负半轴是大陆, x轴是海岸线;

然后在海上有很多海岛, 需要雷达监控, 现在的问题是, 至少需要多少个雷达, 能全部覆盖这些海岛?

 

 

分析：可以化解成区域选最少点问题，注意装换ok

 

代码：

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;

struct node
{
    double x, y;
    bool operator <(node &b)const
    {
        return y<b.y || (y == b.y&&x>b.x);
    }
}point[1000+10];
int main()
{
    int n, d;
    double x, y;
    int kase = 0;
    while (cin >> n >> d&&n != 0 && d != 0)
    {
        kase++;
        int ok = 1;
        int cnt = 0; 
        for (int i = 0; i < n; i++)
        {
            scanf("%lf%lf", &x, &y);
            if (y>d){
                ok = 0;
            }
            else{
                point[i].x = x - sqrt(d*d - y*y);
                point[i].y = x + sqrt(d*d - y*y);
            }
            
        }
        getchar();

        sort(point, point + n);
        double end;
        for (int i = 0; i < n; i++)
        {
            if (i == 0){
                end = point[i].y;
                continue;
            }
            if (end < point[i].x){
                end = point[i].y;
                cnt++;
            }

        }
        if (ok == 0){
            printf("Case %d: -1\n", kase);
        }
        else{
            printf("Case %d: %d\n",kase, cnt + 1);
        }

    }
    

    return 0;
}