POJ 1328 Radar Installation 贪心

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 48402		Accepted: 10768

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：

  
  
  
  

   
   
   
   有一个平面坐标，当前假设 x 轴上方为大海，在海上有 n 座岛屿，需要在海岸线上（ x 轴）建造一个个雷达去覆盖海上的岛屿，而且雷达都是统一的，具有相同的覆盖面积，其半径为 d，要求输出能覆盖所有岛屿的最小雷达数，未满足要求时输出 -1 。
   
   
   
   

  
  
  
  
思路：

  
  
  
  

   
   
   
   想法很简单，就是贪心的思路，一个圆尽量覆盖更多的点（岛屿）。
   
   
   
   
只需要将点的坐标按 x 进行排序，先处理左边的点，依次往右边处理。
   
   
   
   
有两个地方需要注意一下：
   
   
   
   
1、如下图：
   
   
   
   

   
   
   
  
  
  
  

#include <stdio.h>
#include <algorithm>
#include <math.h>

using namespace std;

// 岛屿对象
struct coordinate 
{
    // 坐标、与 X 轴相交线段的起/终点
    double x, y;
    double s, e;
    bool operator<(const coordinate &o) const {
        return e < o.e;
    }
} island[1001];

int solve(int n, int d) 
{
    // 计算相交线段起/终点
    for (int i = 0; i < n; ++i) 
	{
        double t = sqrt(d - island[i].y * island[i].y);
        island[i].s = island[i].x - t;//左侧 
        island[i].e = island[i].x + t;//右侧 
    }

    // 依线段终点排序
    sort(island, island + n);

    // 最少雷达数量（特殊情况已在输入时排除掉，
    // 程序执行到这里时，至少存在一个合法岛屿，
    // 故 ans 初始为 1）
    int ans = 1;
    double r = island[0].e;
    // 从第二座岛屿开始扫描，如前一雷达设置点
    // 覆盖不到，则雷达数加 1
    for (int i = 1; i < n; ++i) {
        if (island[i].s > r) {
            ++ans;
            r = island[i].e;
        }
    }
    return ans;
}

int main() 
{
    int c = 1;
    int n, d;
    while (scanf("%d %d", &n, &d) && (n || d)) 
	{
        bool np = false;

        for (int i = 0; i < n; ++i) 
		{
            scanf("%lf %lf", &island[i].x, &island[i].y);
            if (island[i].y > d) {
                np = true;
            }
        }
        if (np) 
            printf("Case %d: %d\n", c, -1);
		else 
            // 直接传入半径平方，减少重复计算
            printf("Case %d: %d\n", c, solve(n, d * d));
        ++c;
    }
    return 0;
}