acm 1011 雷达数目

1.1011

2.

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. <br> <br>The input is terminated by a line containing pair of zeros <br>

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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1

3.求能够覆盖所有岛屿的最小雷达数目

4.每个小岛对应x轴上的一个区间，在这个区间内的任何一个点放置雷达，则可以覆盖该小岛，区间范围的计算用[x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)];这样，问题转化为已知一定数量的区间，求最小数量的点，使得每个区间内至少存在一个点。每次迭代对于第一个区间， 选择最右边一个点， 因为它可以让较多区间得到满足， 如果不选择第一个区间最右一个点（选择前面的点）， 那么把它换成最右的点之后， 以前得到满足的区间， 现在仍然得到满足， 所以第一个区间的最右一个点为贪心选择， 选择该点之后， 将得到满足的区间删掉， 进行下一步迭代， 直到结束（参考杨世诚）

5.

#include<iostream>

#include<algorithm>

#include<numeric>

#include<math.h>

using namespacestd;

struct str

{

    double l;

    double r;

};

bool cmp(conststr&a,const str&b)

{

    if(a.l<=b.l) return true;

    return false;

}

int main()

{

    str r[1000];

    int n,d,i,l;

    int x,y,t,k,e=1;

    double 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)

            {

                r[i].l=x-sqrt(d*d-y*y);

                r[i].r=x+sqrt(d*d-y*y);

            }

            else l=0;

        }

        if(!l)

        {

            cout<<"Case"<<e++<<": -1"<<endl;

            continue;

        }

        sort(r,r+n,cmp);

        a=r[0].r;

        for(k=1;k<n;k++)

        {

            if(r[k].l>a)

            {

                t++;

                a=r[k].r;

            }

            else if(r[k].r<=a)

                {

                    a=r[k].r;

                }

        }

        cout<<"Case"<<e<<": "<<t<<endl;

        e++;

    }

    return 0;

}