poj 1328

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 72902		Accepted: 16269

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

题意：假设海岸线是一条无限延伸的直线。陆地在海岸线的一侧，而海洋在另一侧。每一个小的岛屿是海洋上的一个点。雷达坐落于海岸线上，只能覆盖d距离，所以如果小岛能够被覆盖到的话，它们之间的距离最多为d。

题目要求计算出能够覆盖给出的所有岛屿的最少雷达数目。

我们假设岛屿i它的x坐标为island[i][0]，而y坐标为island[i][1]，那么有以下几种情况是invalide的，即输出-1的情况：

1.island[i][1]<0

2.abs(island[i][1])<d

3.d<0

其他的情况，应该就是正常情况，进入计算最小雷达数目。



如上图，红色的点为岛屿，那么能够覆盖到此岛屿的雷达所在的区间，应该就是以该岛屿为圆心的圆与x轴交点所在的区间。

这样，我们就可以计算出所有岛屿的雷达所在的区间，得到一个区间数组。

我们将这个数组按照区间左部分进行排序，那么重叠部分就表明这些岛屿的雷达可以共用一个。从而计算出最终解。

 #include
<stdio.h>
#include <string.h>
#include <queue>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
typedef long long LL;
#define max(a,b)(a>b?a:b)
#define min(a,b)(a<b?a:b)
#define N 1005
#define oo 0x3f3f3f3f
using namespace std;

struct point
{
    double l,r;
}p[1100];

int cmp(point a,point b)
{
    return a.l < b.l ;
}

int main()
{
    int n,d,x,y,ca=1,ans,i;
    double s;
    while(scanf("%d%d",&n,&d),n+d)
    {
        int f=0;
        for(i=0;i<n;i++)
        {
            scanf("%d%d",&x,&y);
            if(y>d )
            {
                f=1;
            }
            p[i].l=x-sqrt(1.0*d*d-1.0*y*y);
            p[i].r=x+sqrt(1.0*d*d-1.0*y*y);
        }
        if(f==1)
        {
            printf("Case %d: -1\n",ca++);
            continue;
        }
        else
        {
            sort(p,p+n,cmp);
            ans=1;
            s=p[0].r;
            for(i=1;i<n;i++)
            {
                if(s<p[i].l)
                {
                    s=p[i].r;
                    ans++;
                }
                else if(s>p[i].r)
                {
                    s=p[i].r;
                }
            }
            printf("Case %d: %d\n",ca++,ans);
        }
    }
    return 0;
}