标签：算法 algorithm iostream c++ struct
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
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
struct Island
{
double left;
double right;
}island[1100];
bool cmp(Island a,Island b)
{
return a.left<b.left;
}
int main()
{
int n,d,i,j,t,temp=0;
double x,y,fuck;
while(~scanf("%d%d",&n,&d),n+d)
{
int m=0;
temp++;
for(i=0;i<n;i++)
{
scanf("%lf %lf",&x,&y);
if(y>d||d<0)
{
m=1;
}
island[i].left=x-(sqrt(d*d-y*y));
island[i].right=x+(sqrt(d*d-y*y));
}
if(m)
{
printf("Case %d: -1\n",temp);
continue;
}
sort(island,island+n,cmp);
t=1;
fuck=island[0].right;
for(i=1;i<n;i++)
{
if(island[i].right<=fuck)
{
fuck=island[i].right;
}
else if(island[i].left>fuck)
{
t++;
fuck=island[i].right;
}
}
printf("Case %d: %d\n",temp,t);
}
return 0;
}
POJ-1328 Radar Installation-贪心算法
标签：算法 algorithm iostream c++ struct
原文地址：http://blog.csdn.net/holyang_1013197377/article/details/41145889
