贪心 + 计算几何 --- Radar Installation

Radar Installation

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

 

【题目来源】

Beijing 2002

http://poj.org/problem?id=1328

【解题思路】

以每个岛的坐标为圆心画圆，会与x轴有2个交点，那么这2个点就是能覆盖该岛的雷达x 坐标区间，问题就转变成对一组区间，找最少数目的点，使得所有区间中都有一点。把包含某区间的区间删掉（如果一个点使得子区间得到满足， 那么该区间也将得到满足），这样所有区间的终止位置严格递增。

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婪选择， 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。

ac代码：

 

#include<iostream>

#include<cstdio>

#include<cmath>

#include<algorithm>

using namespace std;

#define MAX 1010

struct Node

{

    float x,y;

    float l,r;

    bool vis;

};

Node node[MAX];

int n;

float r;

bool cmp(Node a,Node b)

{

    return a.r<b.r;

}

int main()

{

//    freopen("in.txt","r",stdin);

    int kase=1;

    while(cin>>n>>r)

    {

        if(n==0&&r==0)

            break;

        int i,j;

        int cnt=0;

        if(r<=0)

            cnt=-1;

        for(i=0;i<n;i++)

        {

            scanf("%f%f",&node[i].x,&node[i].y);  //血的教训，这儿用cout绝对超时

            if(node[i].y>r)

                cnt=-1;

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

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

        }

        printf("Case %d: ",kase++);

        if(cnt==-1)

            {cout<<"-1"<<endl;continue;}

        for(i=0;i<n;i++)

            node[i].vis=false;

        sort(node,node+n,cmp);

        bool flag;

        for(i=0;i<n&&cnt>=0;i++)

        {

            if(!node[i].vis)

            for(j=0;j<n;j++)

            {

                if(!node[j].vis)

                {

                    if(node[j].l<=node[i].r)

                    {

                        node[j].vis=true;

                        flag=true;

                    }

                    else break;

                }

            }

            if(flag)

                cnt++;

            flag=0;

        }

        cout<<cnt<<endl;

    }

    return 0;

}