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Radar Installation
| Time Limit: 1000MS | | Memory Limit: 10000K |
| Total Submissions: 96177 | | Accepted: 21378 |
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
Source
可将点的坐标转为在x轴上对应的位置，然后根据右端对应的值进行排序判断。
#include<cstdio>#include<algorithm>#include<cmath>using namespace std;struct pos{ double left,right;};bool flag;bool compare(pos a, pos b){ return a.right < b.right;}int main(){ int n; double d; int m = 1; pos a[1005]; while (~scanf("%d %lf", &n, &d)) { if (!n || !d) { break; } int sum = 1; flag = true; double x, y; for (int i = 0; i<n; i++) { scanf("%lf %lf", &x, &y); if (!flag) { continue; } if (y > d) { flag = false; continue; } a[i].left = x - sqrt(d*d - y*y); a[i].right = x + sqrt(d*d - y*y); } if (!flag) { printf("Case %d: -1\n",m); m++; continue; } sort(a, a + n,compare); double temp = a[0].right; for (int i = 1; i < n; i++) { if (temp<a[i].left) { sum++; temp = a[i].right; } } printf("Case %d: %d\n", m,sum); m++; } return 0;}
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