Radar Installation
Time Limit:1000MS Memory Limit:10000K Total Submit:2704 Accepted:564
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
诈一看这个题目,或许就有了写想法.觉得不是无从下手,之前思路错了一次.之后看了网上的解题报告,才发现自己错了.
开始错的: 先把所有的点X点从大到小排一下. 再早左极限 再与s=2*(sqrt(d*d-y*y)).比较,错的原因就是如果下一个点,甚至下下个点的左极限在上个点的左极限之内的话,就会出错!
后来改了一点点:先把所有的左极限找出来,再来排左极限,当然 相应的s也要跟随排过去(这样就可以用数组结构体,只是当时用了快排了,就添加了一下,代码也就麻烦了一点.当然数据不多也是一个原因.) 废话不多说,直接贴代码!
#include <stdio.h>
#include <math.h>
void quick_sort(double b[],double a[],int l,int r)//这个快排就是来排左极限的。
{
if(l<r)
{ int i,j ,k,t;
double x,y;
i=l;j=r;x=b[l],y=a[l];
while(i<j)
{
while(i<j&&b[j]>=x)
j--;
if(i<j)
{k=i;t=j;
b[i++]=b[j];
a[k++]=a[t];
}
while(i<j&&b[i]<x)
i++;
if(i<j)
{k=i;t=j;
b[j--]=b[i];
a[t--]=a[k];
}
}
b[i]=x;
a[i]=y;
quick_sort(b,a,l,i-1);
quick_sort(b,a,i+1,r);
}
}
int main()
{
int n,i,j,z,sum,temp=1;
double d,t,p,q;
double a[1500],b[1500],k[1500],s[1500];
while(~scanf("%d %lf",&n,&d))
{ z=0;
sum=0;
if(n==0&&d==0)
break;
for(i=1;i<=n;i++)
{
scanf("%lf%lf",&a[i],&b[i]);
if(b[i]>d)
{
z=1;
}
}
for(i=1;i<=n;i++)
{
k[i]=a[i]-sqrt(d*d-b[i]*b[i]);
s[i]=sqrt(d*d-b[i]*b[i]);
s[i]=2*s[i];
}
quick_sort(k,s,1,n);
for(i=n;i>=1;i--)//这个循环就是来和s比较的。
{ p=k[i];
for(j=i-1;j>0;j--)
{
if(p-k[j]>s[j])//如果两个左极限之间的距离比s还大的话，说明要加一个雷达！
{
sum++;
break;//确定要了之后就从下一个开始判定了.直至最后一个i.
}
i--;
}
if(i==1)//判断最后一次还要不要再装一个雷达.
sum++;//雷达数.
}
if(z==1)
printf("Case %d: -1\n",temp);
else
printf("Case %d: %d\n",temp,sum);
temp++;
}
return 0;
}