SDAU 贪心专题 11 雷达

1：问题描述
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

2：大致题意

x轴的上方是海，下方是陆地。海中有小岛，现在给你小岛的坐标，要在x轴上建立雷达，让雷达覆盖住小岛。求出需要最少的雷达数。

3：思路

我们可以设小岛的坐标为（x，y）。圆心为（a，0）。雷达范围的半径为d，那么我们就可以得到表达式：
（x-a）*（x-a）+y*y=d*d。
可以得到：
x-sqrt(d*d-y*y)小于a
x+sqrt(d*d-y*y)大于a
这里需要注意的是 有可能会出现y>d的情况。要特殊考虑。
得到a的区间之后按照左端点从小到大排列。再根据重叠情况判断最少需要的雷达数即可。

4：感想

这个题的原始表达式好像和豆豆不太一样，不过最终的结果是一样的哎。
找到关系式这个题就挺容易解决的。
5：ac代码

#include <cstdio>#include<iostream>#include<stdio.h>#include<vector>#include<algorithm>#include<numeric>#include<math.h>#include<string.h>#include<map>#include<set>#include<vector>using namespace std;struct dao{    double left;    double right;};bool cmp(const dao &a,const dao &b){    if(a.left<=b.left) return true;    return false;}int main(){    dao ww[1000];    int n,d,i,l;    int x,y,t,k,e=1;    double L,R,a;    while(cin>>n>>d&&n!=0||d!=0)    {        t=1;l=1;        for(i=0;i<n;i++)        {            cin>>x>>y;            if(d>=y&&l)            {                ww[i].left=x-sqrt(d*d-y*y);                ww[i].right=x+sqrt(d*d-y*y);            }            else l=0;        }        if(!l)        {            cout<<"Case "<<e++<<": -1"<<endl;            continue;        }        sort(ww,ww+n,cmp);        R=ww[0].right;        for(k=1;k<n;k++)        {            if(ww[k].left>R)            {                t++;                R=ww[k].right;            }            else if(ww[k].right<=R)                {                    R=ww[k].right;                }        }        cout<<"Case "<<e<<": "<<t<<endl;        e++;    }    return 0;}