POJ 1328：Radar Installation（贪心）

Radar Installation

Time limit:1000 ms Memory limit:10000 kB OS:Linux

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Problem 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

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题意：

X轴以上是海，以下是岸，海上有许多岛屿，现在要在岸边安装雷达，给出雷达能覆盖的半径，问要覆盖所有岛屿最少需要多少雷达。

解题思路：

（从左到右安防雷达）P为岛，根据每个岛的坐标都能算出其右边与其最远的能覆盖到它的雷达的x坐标：

x=xp+R2−y2p−−−−−−√

根据这个x坐标来贪心，如果上一个雷达能覆盖到当前的岛，继续使用上一个雷达，如果不能，就用上面的公式计算出一个新的雷达坐标，总数+1。

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Code：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>

#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;

const int maxn=1000+5;

struct Node
{
    double x,y;
} a[maxn];

int n,d;

bool cmp(Node a,Node b)
{
    double len1= a.x+sqrt(d*d-a.y*a.y);
    double len2=b.x+sqrt(d*d-b.y*b.y);
    return len1<len2;
}

double dis(Node a,Node b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

int main()
{
    int ca=1;
    while(scanf("%d%d",&n,&d)!=EOF)
    {
        if(n==0&&d==0)
            break;
        int flag=1;
        for(int i=0; i<n; i++)
        {

            scanf("%lf%lf",&a[i].x,&a[i].y);
            if(a[i].y>d)
                flag=0;
        }

        printf("Case %d: ",ca++);
        if(flag==0)
        {
            printf("-1\n");
            continue;
        }

        sort(a,a+n,cmp);
        int cnt=1;
        Node last;
        last.y=0;
        last.x=a[0].x+sqrt(d*d-a[0].y*a[0].y);

        for(int i=1; i<n; i++)
        {
            if(dis(last,a[i])>d)
            {
                cnt++;
                last.x=a[i].x+sqrt(d*d-a[i].y*a[i].y);
            }
        }

        printf("%d\n",cnt);

    }
    return 0;
}