POJ 1328 Radar Installation（贪心）


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

x轴代表海岸线，上边代表海洋，下边代表陆地，在海中有岛屿，给出坐标（x，y），陆地有雷达，给出其覆盖半径d，问至少需要几个雷达才能覆盖所有的岛。如果没有解决方案就输出-1

思路:

我们找一个岛屿能被侦测到的极限范围, 当岛屿到雷达的距离等于d时， 雷达可以位于岛屿的左侧也可以位于雷达的右侧。而这就可以分别确定雷达相对于岛屿的最左坐标和最右坐标。

最左为：x - sqrt(d*d-y*y); 最右为：x+ sqrt(d*d-y*y); 每个岛屿都有这样的最左和最右可被侦测坐标。

根据贪心的思想，每次都应该将最右坐标作为衡量标准。
假定当前的岛屿为cur，当前的下一个为next。

1.如果next的最左坐标比cur的最右坐标都大的话，只能再设一个雷达来侦测next了,然后将next的最右坐标作为比较量。

2.如果next的最左坐标比cur的最右坐标小，这时会有两种情况。

A.next最右 < cur最右

B.next最右 >= cur最右
对于A情况,也就等价于next包含于cur, 这样就应该把next的右作为衡量标准了

对于B情况,我们可以直接侦测到next了, 可以找next的next了.

在第二种情况中为什么我们每次都取两个点右坐标的最小值呢，因为半径R一样，那图就有高矮胖瘦之形，若取右坐标的最大值的话，那右坐标小的那个 点就有可能够不到

#include <iostream>

#include <string.h>

#include <math.h>

#include <stdlib.h>

using namespace std;

struct node

{

double xL,xR;

}p[1010];

int cmp(const void*a,const void*b)

{

return ((node*)a)->xL>((node*)b)->xL?1:-1;

}

int main()

{

int n,d,cnt=1;

while(cin>>n>>d&&(n||d))

{

int flag=1;

if(d<=0)

flag=0;

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

{

int x,y;

cin>>x>>y;

if(y>d)

flag=0;

double t=sqrt(1.0*(d*d-y*y));

p[i].xL=x-t;

p[i].xR=x+t;

}

if(flag)

{

qsort(p,n,sizeof(p[0]),cmp);

int sum=1;

double high=p[0].xR;

for(int i=1;i<n;i++)

{

if(p[i].xL>high)

{

sum++;

high=p[i].xR;

}

else

high=high<p[i].xR?high:p[i].xR;

}

cout<<"Case "<<cnt++<<": "<<sum<<endl;

}

else

cout<<"Case "<<cnt++<<": -1"<<endl;

}

return 0；

}