Description
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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<tex2html_verbatim_mark> distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d<tex2html_verbatim_mark> .
We use Cartesian coordinate system, defining the coasting is the x<tex2html_verbatim_mark> -axis. The sea side is above x<tex2html_verbatim_mark> -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<tex2html_verbatim_mark> - y<tex2html_verbatim_mark>coordinates.
Input
The input consists of several test cases. The first line of each case contains two integers n<tex2html_verbatim_mark>(1n
1000)<tex2html_verbatim_mark> and d<tex2html_verbatim_mark> , where n<tex2html_verbatim_mark> is the number of islands in the sea and d<tex2html_verbatim_mark> is the distance of coverage of the radar installation. This is followed by n<tex2html_verbatim_mark> 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 2 1 2 -3 1 2 1 1 2 0 2 0 0
Sample Output
Case 1: 2 Case 2: 1
解题思路:
这个题目的大意为有一个坐标轴 在X轴上方是海 下方是陆地 X轴是海岸线,海上有N个小岛,现在要在海岸线上安装雷达,雷达的覆盖范围是一个以R为半径的圆,请用最少的雷达覆盖所有的小岛;当无法覆盖时 输出-1。我们可以算出 每个小岛能被覆盖的雷达的圆心,即以小岛为圆心 R为半径 作圆,该圆与X轴的交点:左交点为x-sqrt(R*R-y*y); 右交点为x+sqrt(R*R-y*y);按照 左交点 排序,如果i点的左交点在 当前雷达的右边 则需安装一个新雷达,更新雷达否则 如果 i点的右交点也在当前雷达的左边 则把当前雷达的圆心更新为该点的右交点
程序代码:
#include <iostream> #include <algorithm> #include <cmath> using namespace std; const int N = 1001; struct node { double l,r; }a[N]; bool fun(node a, node b) { return a.l<b.l; } int main() { int n,R,i,Case=0; double x,y,t; while(scanf("%d %d", &n, &R)&&n&&R) { int v=1; for(i=0; i<n; i++) { cin>>x>>y; a[i].l = x - sqrt(R*R-y*y); a[i].r = x + sqrt(R*R-y*y); if(y>R || R<=0 || y<0) v=-1; } sort(a, a+n,fun); t = a[0].r; for(i=1; i<n && v!=-1; i++) if(a[i].l>t) { v++; t = a[i].r; } else if(a[i].r<t) t=a[i].r; printf("Case %d: %d ",++Case, v); } return 0; }