POJ 1328 Radar Installation（贪心）

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 30071		Accepted: 6607

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

解题报告：求出每个点的雷达所能安放的范围，再利用经典的活动安排的贪心算法即可

代码如下：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cmath>
using namespace std;
const int N = 1010;
struct point
{
    double x;
    double y;
}p[N]; //点的信息
struct node
{
    double begin;
    double end;
}a[N]; //雷达所能安放的位置范围
int visit[N];
int cmp(const void *a, const void *b)//排序函数
{
    return (*(struct node *)a).end > (*(struct node *)b).end ? 1 : -1;
}
double Max(double x, double y)
{
    return x > y ? x : y;
}
int main()
{
    int n, d, i, j, ans, count = 0;
    double max, len;
    while (scanf("%d%d", &n, &d) != EOF && n && d)
    {
        max = 0;
        memset(p, 0, sizeof(p));
        for (i = 0; i < n; ++i)
        {
            scanf("%lf%lf", &p[i].x, &p[i].y);
            max = Max(max, p[i].y);//找出纵坐标的最大，以便比较和雷达扫描半径的大小
        }
        getchar();
        getchar();
        count ++;
        if (max > d || d < 0)
        {
            printf("Case %d: -1\n", count);
        }
        else
        {
            memset(a, 0, sizeof(a));
            memset(visit, 0, sizeof(visit));
            for (i = 0; i < n; ++i)
            {
                len = sqrt(1.0 * d * d - 1.0 * p[i].y * p[i].y);
                a[i].begin = p[i].x - len;
                a[i].end = p[i].x + len;
            }
            qsort(a, n, sizeof(a[0]), cmp);
            ans = 0;
            for (i = 0; i < n; ++i)
            {
                if (!visit[i])
                {
                    visit[i] = 1;
                    for (j = 0; j < n; ++j)//使用的是经典的活动安排的贪心算法
                    {
                        if (!visit[j] && a[j].begin <= a[i].end)
                        {
                            visit[j] = 1;
                        }
                    }
                    ans ++;
                }
            }
            printf("Case %d: %d\n", count, ans);
        }
    }
    return 0;
}