poj1328 Radar Installation

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 64162 Accepted: 14424

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

 

嗯，题目大意就是说，海岸边装有雷达，岸边的小岛要想能在雷达的监测范围内，雷达在海岸上即x轴上，x轴上方为海洋有小岛，问，所需的雷达个数的最小值，给出小岛坐标以及雷达范围，以小岛为中心，雷达监测范围为半径，利用勾股定理计算区间上下限，再进行排序，区间上限从小到大排，在累加不交叉区间的个数即可

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;struct node {double a,b;}p[1010];bool cmp(node n,node m){if(n.b!=m.b)return n.b<m.b;return n.a<m.a;}int main(){int n,m,i,flag,cnt=0;double x,y;while(scanf("%d %d",&n,&m)&&(n||m)){cnt++;flag=0;for(i=0;i<n;++i){scanf("%lf%lf",&x,&y);if(y>m)flag=1;p[i].a=x-sqrt(m*m*1.0-y*y);p[i].b=x+sqrt(m*m*1.0-y*y);}if(flag){printf("Case %d: -1\n",cnt);continue;}sort(p,p+n,cmp);int s=1;int k=0;for(i=1;i<n;++i){if(p[i].a>p[k].b){s++;k=i;}} printf("Case %d: %d\n",cnt,s);}return 0;}