ACM POJ 1328Radar Installation
Radar Installation
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 27114 | Accepted: 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
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.
![](http://poj.org/images/1328_1.jpg)
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 2 1 2 -3 1 2 1 1 2 0 2 0 0
Sample Output
Case 1: 2 Case 2: 1
Source
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;
}
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