Radar Installation
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 52823 | Accepted: 11883 |
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
思路 ：在X轴上肯定有个区域，雷达在这个区域内可以把某个小岛覆盖住；然后把这些区域按左区间排序，然后贪心的求解就行了。
这些仔细考率都能过的，因为一个变量搞成int型竟然错了好久！
#include"stdio.h"
#include"string.h"
#include"math.h"
#include"iostream"
#include"algorithm"
using namespace std;
#define N 1005
struct node
{
double l,r;
}p[N];
double x[N],y[N];
bool cmp(node a,node b)
{
return a.l<b.l;
}
int main()
{
int i,n,d,cnt=1,flag;
while(scanf("%d%d",&n,&d),n||d)
{
flag=0;
for(i=0;i<n;i++)
{
scanf("%lf%lf",&x[i],&y[i]);
if(flag)
continue;
if(d<y[i])
flag=1;
double t=sqrt(d*d-y[i]*y[i]);
p[i].l=x[i]-t;
p[i].r=x[i]+t;
}
printf("Case %d: ",cnt++);
if(flag)
{
printf("-1\n");
continue;
}
sort(p,p+n,cmp);
int t=1;
double rr;
rr=p[0].r;
for(i=1;i<n;i++)
{
if(p[i].l>rr) //该区间不能被雷达覆盖住
{
rr=p[i].r;
t++;
}
if(p[i].r<rr) //可以被当前雷达作用，
{
rr=p[i].r;
}
}
printf("%d\n",t);
}
return 0;
}