Radar Installation(贪心)

Language:
Radar Installation
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 88059 Accepted: 19733
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

代码:

//就是把每个岛屿的圆心给求出来,前提得知//道的是,只有当岛屿在圆上的时候,它才有更多的空间去包含其他的岛屿//就用sqrt(d*d-arr[i].y(arr[i].y)),得到圆心后就把圆心进行排序,再把岛屿和圆心进//行比较,如果可以包含在内,则继续添加岛屿,如果出现无法包含到一个//圆心时,则需要再开一个圆心,以此类推#include <iostream>#include <algorithm>#include <cmath>using namespace std;struct island{    float x;//岛屿的横坐标    float y;//岛屿的纵坐标    float r;//圆心的横坐标};island arr[1000+10];//将圆心按从小到大进行排序bool compared(island a,island b){    return a.r<b.r;}int main(){    int n,d,ans=1,cns=0;    int flag=0;    while(cin>>n>>d)    {        cns++;        ans=1;        flag=0;        if(n==0 && d ==0)            break;        for(int i=0;i<n;i++)        {            cin >> arr[i].x >> arr[i].y;            arr[i].r=arr[i].x+sqrt(d*d-arr[i].y*arr[i].y);//计算每个岛屿的圆心的横坐标            if(arr[i].y>d)                flag=1;        }        if(flag==1 || d<0)        {            cout << "Case " << cns << ": -1" <<endl;            continue;        }        //给圆心进行排序        sort(arr,arr+n,compared);        float temp=arr[0].r;        float a=0,b=0,c=0;        for(int i=0;i<n;i++)        {            a=arr[i].x-temp;            b=arr[i].y;            if((a*a+b*b)>d*d)//该岛屿不在该圆心的范围内时,增加一个圆心(雷达),即该岛屿的圆心(雷达)            {                ans++;                temp=arr[i].r;            }        }        cout << "Case " << cns << ": " << ans << endl;    }    return 0;}