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

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=;

    while(cin>>n>>r)

    {

        if(n==&&r==)

            break;

        int i,j;

        int cnt=;

        if(r<=)

            cnt=-;

        for(i=;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=-;

            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==-)

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

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

            node[i].vis=false;

        sort(node,node+n,cmp);

        bool flag;

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

        {

            if(!node[i].vis)

            for(j=;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=;

        }

        cout<<cnt<<endl;

    }

    return ;

}

巴特西

贪心 + 计算几何 --- Radar Installation

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