POJ 1328 Radar Installation

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 43793 Accepted: 9704

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

这个题据说是贪心题，但是网上的解题报告看来看去，觉得更像是区间覆盖，然后自己就用这样的思路写了一下，结果就AC了，具体就是先将平面上的问题化成数轴上的问题，将平面每个点坐标得到后先进行判断，如果纵坐标大于雷达半径，则输出-1结束，如果不是，运用勾股定理计算在X轴修建雷达可以辐射的区间，用结构体point的两个值left以及right存储区间的左值以及右值，按左值从小到大对结构体point进行排序，将第一个雷达放置在最左边第一个区间内，并且判定位置放到第一个区间的右值，判断下一个点的左值是否在判定点左面，如果在，则此雷达可以同时扫描到该点，继续判定右值是否在该判定点左面，如果在，将判定点变为该点的右值，如果左值不在判定点左面，则需添加雷达，雷达数量加1，并且将判定点变为该点右值，直到结束所有点的判定后输出当前雷达数量。

下面是AC的代码：

#include<cstdio>#include<cmath>#include<iostream>using namespace std;struct st{double left;double right;}point[1005];int x[1005],y[1005];void sort(int n){   int i,j,k;   st temp;    for (i=0;i<n;i++)        {        k=i;        for (j=i;j<n;j++) if (point[j].left<point[k].left) k=j;        temp=point[i];point[i]=point[k];point[k]=temp;        }}int main(){int i,time=1,num,flag,n,d;double r,temp;while(1){num=1;flag=1;scanf("%d%d",&n,&d);if(n==0&&d==0)break;for(i=0;i<n;i++){scanf("%d%d",&x[i],&y[i]);if(y[i]>d)flag=0;}if(flag==0){printf("Case %d: -1\n",time);time++;}else{for(i=0;i<n;i++){temp=sqrt(d*d-y[i]*y[i]);    point[i].left=(double)x[i]-temp;    point[i].right=(double)x[i]+temp;}sort(n);r=point[0].right;for(i=1;i<n;i++){if(point[i].left<=r){if(point[i].right<=r)r=point[i].right;}else{r=point[i].right;num++;}}printf("Case %d: %d\n",time,num);time++;}}return 0;}

	
				
	
					   
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