[ACM] POJ 1328 Radar Installation (贪心，区间选点问题）

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

Source

Beijing 2002

解题思路：

题意为，在一个坐标系中，x中代表海岸，x轴以上有n个点，代表着n个岛屿，在x轴上建雷达，一直雷达的覆盖范围为半径为d的圆，求在x轴上最少建多少个雷达，才能把全部的岛屿覆盖起来，如果不能覆盖，输出-1.

区间选点问题为  给定 n个闭区间,求在里面选择最少的点使得每个区间里面都包含至少一个点（一个点可以在不同的区间)。   比如下图。

贪心策略为  把n个区间先按照右端点从小到大进行排序，如果端点相同，按照左端点从大到小进行排序。首先选择第一个区间的右端点。如果以后的区间左端点大于当前选择的端点值时，使得当前选择的端点值改变为这个以后的区间的右端点值。 比如 当前temp是 23  遇到一个区间左端点 25 右端点 27，让temp=27.（选择了一个新点）

回到本题上来，要求建造最短的雷达。  首先对n个岛屿进行画圆，与x轴有两个交点，这样每个岛屿都对应了一个区间（只要雷达建在这个区间内，该岛屿就一定能被覆盖到）。圆的方程(x－a)2+(y－b)2=r2 ,在本题中，b始终是0，因为在x轴上。判断无解的条件为r<y
。我们得到了n个区间，这样问题也就转化为了在这n个区间里面选择尽可能少的点，使得这n个区间中每个区间都至少有一个点。 按照上面的方法做就可以了。

参考：

http://blog.csdn.net/dgq8211/article/details/7534776

代码：

#include <iostream>
#include <string.h>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn=1010;

struct N
{
    double l,r;
}interval[maxn];

bool cmp(N a,N b)//排序
{
    if(a.r<b.r)
        return true;
    else if(a.r==b.r)
    {
        if(a.l>b.l)
            return true;
        return false;
    }
    return false;
}

int main()
{
    double x,y;
    int n,d;
    int c=1;
    while(cin>>n>>d&&(n||d))
    {
        bool ok=1;
        for(int i=1;i<=n;i++)
        {
            cin>>x>>y;
            double temp=d*d-y*y;
            if(temp<0||d<0)//这样的话就得判断d是否小于0,如果 temp= d-y 这样写的话，就不用判断d是否小于0，坑啊！！
                ok=0;
            else if(ok)
            {
                interval[i].l=x-sqrt((double)d*d - (double)y*y);
                interval[i].r=x+sqrt((double)d*d - (double)y*y);
            }
        }
        if(!ok)
        {
            cout<<"Case "<<c++<<": "<<-1<<endl;
            continue;
        }
        sort(interval+1,interval+1+n,cmp);
        int cnt=1;
        double temp=interval[1].r;
        for(int i=2;i<=n;i++)
        {
            if(temp<interval[i].l)
            {
                cnt++;
                temp=interval[i].r;
            }

        }
       /* for(int i=2;i<=n;i++)
        {
            if(interval[i].l>temp)
            {
                cnt++;
                temp=interval[i].l;
            }
        }*/   //一开始写成了这个，不对  比如区间 [1,2]  [4,8]  [6,9],正确的应该是选择2，8这两个点，而上面这个则选择了 2,4 6，显然不对
        cout<<"Case "<<c++<<": "<<cnt<<endl;
    }
    return 0;
}

[ACM] POJ 1328 Radar Installation (贪心，区间选点问题）