Radar Installation（贪心）

Radar Installation（贪心）

Radar Installation

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

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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 1 2 1 1 2 0 2 0 0

Sample Output

Case 1: 2Case 2: 1

代码：

#include#include#include#include#includeusing namespace std;struct ac{ double l; double r;}p[1500];bool cmp(ac x,ac y){ return (x.r!=y.r)?(x.ry.l);}int main(){ int n,rad; int d=1; while(~scanf("%d%d",&n,&rad),(n&&rad)) { int x,y; int f=1; for(int i=0;i=y) { p[i].l=x-sqrt((double)rad*rad-(double)y*y); p[i].r=x+sqrt((double)rad*rad-(double)y*y); } else { f=0; } } if(!f) { printf("Case %d: -1\n",d++); continue; } sort(p,p+n,cmp); double End=p[0].r; int cnt=1; for(int i=1;i

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