POJ 1328 Radar Installation

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 36207		Accepted: 8039

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

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这题目在知道贪心的情况下仍然错了很多次，看了看解题报告才知道自己的贪心策略有问题，可以求覆盖每个岛的的区间，对区间按左端点进行排序，这样这个问题就转化成了区间覆盖问题

#include <iostream>
#include <string.h>
#include <math.h>
#include <stdlib.h>
using namespace std;
class num
{
public:
double x1,x2;
}a[10000];
int cmp(const void *e,const void *f)
{
num *p1=(num *)e;
num *p2=(num *)f;
if(p1->x1>p2->x1)
{
return 1;
}else if(p1->x1<p2->x1)
{
return -1;
}else
{
return 0;
}
}
int main()
{
int i,j,n,m,s,t,x,y;
int r,k;
double b,c,right;
t=1;
while(cin>>n>>r)
{
if(!n&&!n)
{
break;
}
k=0;
for(i=0;i<=n-1;i++)
{
cin>>x>>y;
if(y>r)
{
k=1;
}
if(k==0)
{
b=sqrt((double)(r*r-y*y));
a[i].x1=(double)x-b;
a[i].x2=(double)x+b;
}
}
if(k==1)
{
cout<<"Case "<<t<<":"<<" "<<"-1"<<endl;
t++;
continue;
}
qsort(a,n,sizeof(a[0]),cmp);
s=1;
right=a[0].x2;
for(i=1;i<=n-1;i++)
{
if(a[i].x1-right>0)
{
s+=1;
right=a[i].x2;
}else
{
if(a[i].x2-right<0)
{
right=a[i].x2;
}
}
}
cout<<"Case "<<t<<":"<<" "<<s<<endl;
t++;
}
return 0;
}