区间选点问题

Radar

时间限制：1000 ms  |  内存限制：65535 KB

难度：3

描述

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.

 

输入

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

输出

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.

样例输入

3 21 2-3 12 11 20 20 0

样例输出

Case 1: 2Case 2: 1

来源

Beijing 2002

上传者

ctest

题意：

在x轴上安装雷达，至少需要多少的雷达，能够将所有的岛屿全部覆盖上去。

表面现象不可信，其实就是一个区间选点的问题。对于这道题来说真的是主要还是贪心思想。

这道题就是将所有的岛屿转化成一系列的区间，转化的方式就是。岛屿的横坐标和岛屿纵坐标与雷达半径的之间的关系，构成的一个区间

只要能知道将这个转化成区间问题，就基本上可以说是做出来一大半了。

区间选点的问题：

  【a,b】,按照b的升序，当b相等的时候。按照a的降序这样不断地进行。

再来一个for循环，找出其中需要多安装的雷达就行了。寻找的方式是，固定一点，找不相邻的区间的个数。

代码：

struct node
{
    double lf,rg;
}s[1005];
bool cmp(node a,node b)
{
    if(a.rg==b.rg)
        return a.lf>=b.lf;
    else
        return a.rg<b.rg;
}
int main()
{
    int n,z=1;
    double d;
    while(scanf("%d%lf",&n,&d)&&n&&d)
    {
        int flag=0;
       double x,y,t;
      for(int i=0;i<n;i++)
      {
         scanf("%lf%lf",&x,&y);
         if(fabs(y)>d)
            flag=1;
         t=sqrt(d*d-y*y);
         s[i].lf=x-t;
         s[i].rg=x+t;
      }
      sort(s,s+n,cmp);
      int sum=1;
      double l=s[0].rg;
     // printf("***%lf\n",s[0].rg);
      for(int i=1;i<n;i++)
      {
         // printf("****%lf %lf\n",s[i].lf,s[i].rg);
          if(l>=s[i].lf)
            continue;
          else
          {
              sum++;
              l=s[i].rg;
          }
      }
      if(flag==1)
        printf("-1\n");
      else
      {
          printf("Case %d: %d\n",z++,sum);
      }
    }
}