POJ-----1328

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 75260 Accepted: 16841

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

对于一个小岛，雷达放置在海岸线（即x轴）一个区域内都可以侦测到该岛，就管这个叫覆盖区域吧，求出每个岛的覆盖区域，

就转化成了若干个区间的交集，按左界升序排列，然后就是贪心的处理

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;struct node{double x, y;}s[1010];bool cmp(node a, node b){return a.x < b.x;}int main(){int n, kcase = 1;double a, b, flag, d;while(~scanf("%d%lf", &n, &d), n || d){int ok = 0;for(int i = 0; i < n; i++){scanf("%lf%lf", &a, &b);s[i].x = a - sqrt(d*d - b*b);//求覆盖区域s[i].y = a + sqrt(d*d - b*b);if(b > d || d < 0){ok = 1;}}if(ok){printf("Case %d: -1\n", kcase++);continue;}sort(s, s + n, cmp);flag = s[0].y;int ans = 1;for(int i = 1; i < n; i++){if(s[i].y <= flag){//右界为判断点，更新区域边界flag = s[i].y;continue;}if(s[i].x > flag){//当一个岛的左界已经不再上一个区域，就以这个岛的右界开启新的区域ans++;flag = s[i].y;}}printf("Case %d: %d\n", kcase++, ans);}return 0;}