Radar Installation
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 20000/10000K (Java/Other)
Total Submission(s) : 60 Accepted Submission(s) : 11
Problem 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.
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
Case 2: 1
Source
PKU
今天通过这道题又学习了一下贪心算法的知识，感觉收获很多。
对于每个岛屿的位置，如果y大于d则雷达无法侦测到岛屿，结果为-1；如果y都小于d，将每个岛屿按照x坐标从小到大排列，并算出对于每个岛屿，一个雷达能侦测到这个岛屿时雷达在x轴上的最左和最右的位置，并储存为lx和rx。从最左边的岛屿算起，如果该岛屿（记为a1）右边下一个岛屿（记a2）的lx大于a1的rx，则只能再加一个雷达才能侦测到a2；如果a2的lx小于a1的rx，且a2的rx小于a1的rx则a2的再下一个岛屿（记为a3）的lx必须小于a2的rx才能被原来的雷达检测到，否则要再加一个雷达（即a3必须和a2的rx比较）；如果a2的lx小于a1的rx，且a2的rx大于a1的rx，能侦测到两个岛屿的雷达坐标范围是a2的lx到a1的rx，则a3必须和a1的rx进行比较来判断是否应该再加一个雷达。
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
struct island
{
int x,y;
double rx,lx;
}; int cmp(const island &a,const island &b)
{
if(a.x<b.x)
return 1;
else
return 0;
} int main()
{
int cas=0,n,d;
while(cin>>n>>d&&n+d)
{
cas++;
island is[1001];
bool f=true;
for(int i=0;i<n;i++)
{
cin>>is[i].x>>is[i].y;
if(is[i].y>d)
f=false;
double t=sqrt(d*d-is[i].y*is[i].y);
is[i].lx=is[i].x-t;
is[i].rx=is[i].x+t;
}
if(!f)
{
cout<<"Case "<<cas<<": "<<-1<<endl;
}
else
{
sort(is,is+n,cmp);
/*for(int i=0;i<n;i++)
cout<<is[i].x<<' '<<is[i].rx<<' '<<is[i].lx<<endl;*/
double temp=is[0].rx;
int count=1;
for(int i=1;i<n;i++)
{
if(is[i].lx>temp)
{
count++;
temp=is[i].rx;
}
else if(is[i].rx<temp)
temp=is[i].rx;
}
cout<<"Case "<<cas<<": "<<count<<endl;
}
}
}