POJ 1328 Radar Installation（贪心法）

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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1

题意：用x轴表示海岸，陆地在x轴下方，海在x轴上方。每个小岛都是海中的一个点。雷达只能安装在x轴上，覆盖范围为d。问最少用多少个雷达能把所有的岛屿覆盖。
思路：先通过平面几何的知识计算出各个岛屿的雷达区间（如图），这样子我们就能得到一段段区间，接着按左端点从小到大排序，然后分情况讨论：
一、输出-1的情况：     1.d<0
                                   2.|y|>d
                                   3.y<0
二、讨论区间之间的关系：       1.相离：上个区间的右端点<下个区间的左端点
                                                2.包含：上个区间的右端点>下个区间的右端点
                                                3.相交但不包含：上个区间的右端点>下个区间的左端点
用贪心法的思想，把雷达安装在区间的最右处（右端点），这样雷达能覆盖到下个小岛的机率更高，如果是第一种相离的情况，说明雷达无法覆盖下个小岛，则要多用一个雷达；如果是第二种包含的情况，也说明雷达无法覆盖下个小岛，此时把雷达的位置改为下个区间的右端点（这样两个小岛都能覆盖到）；如果是第三种相交但不包含的情况，说明雷达两个岛都能覆盖，则直接判断下个区间即可。

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;int n;double d;struct pos{double x,y;} a[1005];struct dis{double left,right;} b[1005];bool cmp(dis A,dis B){return A.left<B.left;}int main(){int k=1;while(scanf("%d%lf",&n,&d)&&n&&d){bool flag=false;if(d<0)flag=true;memset(a,0,sizeof(a));memset(b,0,sizeof(b));for(int i=0; i<n; i++){scanf("%lf%lf",&a[i].x,&a[i].y);if(fabs(a[i].y)>d||a[i].y<0)flag=true;double tmp=sqrt(d*d-a[i].y*a[i].y);b[i].left=a[i].x-tmp;b[i].right=a[i].x+tmp;}if(flag){printf("Case %d: -1\n",k++);continue;}sort(b,b+n,cmp);int cnt=1,i=1;double radar=b[0].right;while(i<n){if(b[i].left>radar)//相离 {cnt++;radar=b[i].right;}/*else if(b[i].left<radar&&b[i].right>radar) //相交不包含{}*/else if(b[i].left<radar&&b[i].right<radar)  //包含 {radar=b[i].right;}i++;}printf("Case %d: %d\n",k++,cnt);}}