POJ 1328 Radar Installation （贪心）

2019-11-08 03:09:10

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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 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1

题意

海上有n多岛，在海岸线上（x轴）建一个雷达能覆盖到与它距离不超过d的岛，求覆盖所有岛的最小雷达数。

思路

以岛屿为圆心，d为半径画圆，交x轴于s,e两点，然后以线段末尾为排序标准对所有这样的线段排序，从左至右依次判断，假设雷达放置在一个线段的末端，然后忽略它所能涉及到的岛屿。

AC 代码

#include <iostream>#include<stdio.h>#include<math.h>#include<string.h>#include<algorithm>#include<stdlib.h>#include<queue>using namespace std;struct point{ double x,y; double s,e; bool Operator<(const point &o)const { return e<o.e; }} land[1005];int solve(int n,int d){ for(int i=0; i<n; i++) //计算所有的线段 { double t=sqrt(d-land[i].y*land[i].y); land[i].s=land[i].x-t; land[i].e=land[i].x+t; } sort(land,land+n); //排序 int ans=1; double r=land[0].e; for(int i=1; i<n; i++) { if(land[i].s>r) { ++ans; r=land[i].e; } } return ans;}int main(){ int n,d,cat=1; while(~scanf("%d%d",&n,&d)&&(n||d)) { bool flag=false; for(int i=0; i<n; i++) { scanf("%lf%lf",&land[i].x,&land[i].y); if(land[i].y>d)flag=true; //是否可以检测到该岛屿 } if(flag)printf("Case %d: -1/n", cat++); else printf("Case %d: %d/n", cat++, solve(n,d*d)); } return 0;}

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