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Radar Installation
| Time Limit: 1000MS |
|
Memory Limit: 10000K |
| Total Submissions: 54970 |
|
Accepted: 12381 |
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
题目翻译：给出两个数字n和d，n代表n个P点，d代表圆的半径，然后下面n行，每行一组坐标（x，y），分别代表所有的P点，
求：在x坐标轴上，至少要画几个以d为半径的圆，才能使P点均在圆内。
解题思路：分别以P点为圆心，d为半径画圆，圆分别会在x轴上截取相应的线段，只要线段内有圆心，则即可覆盖P点，因此问题
便转化为求X轴上共有多少独立区间！
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<math.h>
using namespace std;
struct A{
double zuo,you;
};
int cmp(A a,A b){
return a.zuo<b.zuo;
}
int main(){
int t=1,n,i,r,x,y,yes;
A a[1200];
while(scanf("%d%d",&n,&r)&&!(n==0&&r==0)){
yes=0;
for(i=1;i<=n;i++){
scanf("%d%d",&x,&y);
if(r<y) yes=1;
double num=sqrt((double)(r*r-y*y));
a[i].zuo=double(x-num);
a[i].you=double(x+num);
}
if(yes){printf("Case %d: -1\n",t++);continue;}
sort(a+1,a+n+1,cmp);
double k=a[1].you;
int count=1;
for(i=2;i<=n;i++){
if(a[i].zuo>k){
k=a[i].you;
count++;
}
else{
if(a[i].you<k)
k=a[i].you;
}
}
printf("Case %d: %d\n",t++,count);
}
return 0;
}
poj 1328 【Radar Installation】【几何转化、区间覆盖】
标签：des style blog http io ar color os sp
原文地址：http://blog.csdn.net/hpuhjl/article/details/41512623
