贪心算法解决雷达站建站问题

问题描述:

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.

解题:

        因为所有的雷达站都是在X坐标轴上，而所有的岛都是在X轴的上方。因此，若要找出雷达站的数量，首先应该弄清楚雷达站的坐标。

     （1）对于坐标为(x,y）的小岛，若要使位于x轴上的雷达站覆盖到它，则以此小岛为圆心，以雷达覆盖半径为半径画圆，此圆与x轴的交界区间即为雷达的部署区间。即为 ( x-sprt(d*d-y*y) ，x+sprt(d*d-y*y) )。设小岛i的左右区间点分别放在island[i][0],island[i][1]。

  （2）以（1）中的方法找出每个小岛如果被覆盖，雷达站安置的区间。并以所有区间的右区间点从小到大进行排列。即依据island[i][1]进行排列。

  （3）首先将island[0][1]设为第一个雷达站点，并将所有左区间点island[j][0]小于island[0][1]的小岛去掉，其右区间点island[j][1]随之在数组中去掉。

  （4）对于剩下的所有island[k][0]重复（3）的过程，直到没有小岛为止。没重复进行一次（3）的操作，说明雷达站建起一座。

pseudo-code

    i=num(island);

   while(i--)

   {

     island[i][0]=land[i].x-sprt(d*d-land[i].y*land[i].y);

     island[i][1]=land[i].x+sprt(d*d-land[i].y*land[i].y);

    }

   sort(island);//依据island[i][1]进行排序；

   for(k=0;k

   {

        for(j=k+1;j

         {

           if(island[j][0]

            delete island[j];//如果小于即改点已被雷达覆盖;

          }

         radar_count++;

         sort(island);//继续对剩下的点排序；

         找出新排序的第一个点作为新k;

    }

    return radar_count;

complexity of your algorithm：

   由于在for循环里面嵌套一个排序算法，因此时间复杂度为（n^2*logn）