ACM POJ 1328Radar Installation

Radar Installation

Time Limit: 1000MSMemory Limit: 10000K
Total Submissions: 27114Accepted: 5912

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 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2
Case 2: 1

Source

Beijing 2002       贪心:

#include<stdio.h>
#include
<iostream>
#include
<math.h>
#include
<algorithm>//sort所在的库文件,排序用
using namespace std;
const int MAXN=1005;
struct Line
{
double l,r;
}line[MAXN];
//每个岛作半径为d的圆,与x轴所截的线段

bool cmp(Line a,Line b)
{

return a.l<b.l;
}
//按照线段的左端点从小到大排序
int main()
{
//freopen("test.in","r",stdin);
//freopen("test.out","w",stdout);
int n,d;
int i;
int x,y;
bool yes;//确定是不是有解
int icase=1;
while(cin>>n>>d)
{
yes
=true;
int cnt=0;
if(n==0&&d==0)break;
for(i=0;i<n;i++)
{
cin
>>x>>y;
if(yes==false)continue;
if(y>d)yes=false;
else
{
line[i].l
=(double)x-sqrt((double)d*d-y*y);
line[i].r
=(double)x+sqrt((double)d*d-y*y);
}
}
if(yes==false)
{
cout
<<"Case "<<icase++<<": -1"<<endl;
continue;
}
sort(line,line
+n,cmp);
cnt
++;
double now=line[0].r;
for(i=1;i<n;i++)
{

if(line[i].r<now)//这点很重要
now=line[i].r;
else if(now<line[i].l)
{
now
=line[i].r;
cnt
++;
}
}
cout
<<"Case "<<icase++<<": "<<cnt<<endl;

}
return 0;
}

    原文作者:kuangbin
    原文地址: https://www.cnblogs.com/kuangbin/archive/2011/07/30/2121838.html
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
点赞