POJ 1328 Radar Installation（贪心）

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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1

题意：你要在海岸线上布置一些相同的雷达（x轴）（雷达监测范围是一个半径为R的圆），以监测在大海中的一些小岛（y轴正半轴），给出小岛的坐标x，y，以及雷达的半径。问最少需要布置多少雷达

题解：

这道题有点需要认真思考一下，可以这样生成思路，想象一个半圆，从x负半轴的无穷远平移过来，直到它的左边的1/4圆弧与某个节点相重合，删掉被半圆包含的点，如此

重复，直到所有的圆都被删除完。可以这样证明这个贪心策略的正确性：对于会与左边的1/4圆弧与相重合的节点，如果不是在重合的时候最优，那么就是在左边或右边最优

先考虑右边，如果是右边的话，这个节点就不会被圆包含，那么就需要两个圆，如果在左边，覆盖范围不如刚好重合的时候，故此，刚好重合时最优。

代码实现如果按照上述思路来说很困难，需要转换一下，在找那个与某个刚好重合的圆的时候，不可能算圆系方程吧？我们是以小岛为圆，以半径R作圆，与x轴的右交点就是

那个圆的圆心了，这个可以有勾股定理算，我们也并不需要如此麻烦的去删除节点，在以小岛为圆的时候，我们存储下每个小岛与x轴的左右交点，按右端点排序，那么对于两

个小岛来说，如果i小岛的右交点比j小岛的左交点大（i<j），那么根据几何意义（画个图就可以了，比较显然），j小岛是可以被i小岛的圆所包含的

代码：

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<cstdlib>#include<algorithm>const int MAXN=1005;using namespace std;struct node{double x,y;}island[MAXN];struct Node{double left,right;}circle[MAXN];int n;double d,esp=0.000001;double dist(node a,double b){    return sqrt(b*b-a.y*a.y);}bool cmp(Node a,Node b){    return a.right<b.right;}int main(){    int cas=0;    while(1)    {        int ans=0; bool flag=0;        scanf("%d%lf",&n,&d);        if(!n&&!d) return 0;        for(int i=1;i<=n;i++)        {            scanf("%lf",&island[i].x);            scanf("%lf",&island[i].y);            if(island[i].y>d) flag=1;            if(island[i].y<0) flag=1;        }        printf("Case %d: ",++cas);        if(flag)        {            putchar('-'); putchar('1'); putchar(10);            continue;        }        for(int i=1;i<=n;i++)        {            circle[i].left =island[i].x-dist(island[i],d);            circle[i].right=island[i].x+dist(island[i],d);        }        sort(circle+1,circle+n+1,cmp);        for(int i=1;i<=n;)        {            /*            int j; ans++;            for(j=i+1;j<=n&&circle[i].right>=circle[j].left;j++);            i=j;*/            double last=circle[i].right;            ans++;i++; while(i<=n&&circle[i].left<=last)i++;        }        printf("%d\n",ans);    }}