POJ 1328 Radar Installation
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Radar Installation
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 77830 Accepted: 17412
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
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
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解析:题目其实很简单,就是简单的贪心+sort;以给的点为圆心做圆,圆与x轴交于两点,一点是s.start,另一点是s.end,如果与x轴不相交,则输出-1;然后就会得到s的数组s[1-n];然后以s.endsort一遍,就得到了有顺序的一组数;现在就是简单的贪心了;#include <bits/stdc++.h>const double INF=0x3f3f3f3f;using namespace std;struct node{ double start, end;};node s[1500];double cmp(node a, node b)//以s.end升序写的cmp;{ return a.end<b.end;}int main(){ int n, d, i; double x, y; int K=0; while(cin>>n>>d) { if(n==0||d==0) return 0; int flag=1; for(i=0; i<n; i++) { cin>>x>>y; if(y<=d) { s[i].start=x-sqrt(d*d-y*y); s[i].end=x+sqrt(d*d-y*y); } else flag=0; } cout<<"Case "<<++K<<": "; if(flag==0) { cout<<"-1"<<endl; } else { sort(s,s+n,cmp);//快排:s.end升序// for(i=0;i<n;i++)// printf("%.2lf %.2lf\n",s[i].start, s[i].end); int sum=0; double temp = -INF; for(i=0; i<n; i++) { if(temp<s[i].start) { sum++; temp=s[i].end; } } cout<<sum<<endl; } } return 0;}
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