poj 1328 Radar Installation

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 48743		Accepted: 10884

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

Radar Installation

做的poj第一个题，题意就是在海岸上装最少的雷达（扫描半径为d）使其侦测到所有的小岛.

思路：求出小岛可以被扫描到的雷达的安装范围p[i].left,p[i].right,算完后对p排序，right大的在后面，若相同则left小的在后面。排完序后维持now_left就行了

代码：

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iostream>
using namespace std;
struct pp
{
    double left;
    double right;
}p[1005];
double left(int x,int y,int d)
{
    return x-sqrt(d*d-y*y);
}
double right(int x,int y,int d)
{
    return x+sqrt(d*d-y*y);
}
bool cmp(pp p1,pp p2)
{
    if(p1.right<p2.right)
    {
        return 1;
    }
    else
    {
        if(p1.right==p2.right)
            return p1.left>p2.left;
    }
    return 0;
}
int main()
{
    int n,d,Case=1;;
    while(scanf("%d%d",&n,&d)!=EOF)
    {
        if(!n&&!d)break;
        int i,ok=1;;
        for(i=0;i<n;i++)
        {
            int x,y;
            //cout<<"input"<<endl;
            scanf("%d%d",&x,&y);
            if(y>d)
                ok=0;
            if(ok)
            {
                p[i].left=left(x,y,d);
                p[i].right=right(x,y,d);
            }
        }
        //cout<<"ok"<<endl;
        if(!ok)
        {
            printf("Case %d: -1\n",Case++);
            continue;
        }
        sort(p,p+n,cmp);
        int count=1;
        double now_left=p[n-1].left;
        for(i=n-2;i>=0;i--)
        {
            if(p[i].right<now_left)//这个小岛不能被已安装的雷达扫到
            {
                count+=1;
                now_left=p[i].left;//更新left
            }
            else
            {
                now_left=(now_left<p[i].left?p[i].left:now_left);//更新left
            }
        }
        printf("Case %d: %d\n",Case++,count);
    }
    return 0;
}