Radar Installation POJ - 1328

Posted 2020-10-26 Fitz

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

该题题意是为了求出能够覆盖所有岛屿的最小雷达数目

这题一开始根本没有贪心思路，想不到

qu[i].left=x-sqrt(r*r-y*y);
qu[i].right=x+sqrt(r*r-y*y);

转化为一个区间问题。

关于这个的合理性 你们画个图就好了 ，表示不会电脑画图

 

 

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cctype>
using namespace std;
struct node
{
    double left,right;
}qu[1010];
int cmp(node a,node b)
{
    return a.right<b.right;
}
int main() {
    int n,k=1;
    double r;
    while(scanf("%d%lf",&n,&r)!=EOF){
        if (n==0 && r==0 ) break;
        int flag=1;
        double x,y;
        for (int i=0 ;i<n ;i++){
            scanf("%lf%lf",&x,&y);
            if (!flag) continue;
            if (y>r) {
                flag=0;
                continue;
            }
            qu[i].left=x-sqrt(r*r-y*y);
            qu[i].right=x+sqrt(r*r-y*y);
        }
        sort(qu,qu+n,cmp);
        printf("Case %d: ",k++);
        if (!flag) {
            printf("-1\n");
            continue;
        }
        int sum=0;
        double temp=-10000000;
        for (int i=0 ; i<n ;i++){
            if (qu[i].left>temp){
                sum++;
                temp=qu[i].right;
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}