Radar Installation / POJ - 1328

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

题意

多组测试数据
给你n个坐标上的点, n <= 1000
你可以在x轴上放置雷达,每个雷达的辐射半径都是d
现在问你辐射到所有的点至少需要几个雷达,如果辐射不到所有的点输出-1

题解

转化为区间问题
已知一点，可求出对应的最短距离的雷达（在X轴上）的位置， 此时雷达覆盖范围在X轴上的左右端点就是区间的左右端点「需要数学公式」。

https://www.cnblogs.com/Little-Turtle--QJY/p/12390466.html
「数学原理！！啦」

代码

#include<bits/stdc++.h>
using namespace std;

struct note{
	double l, r;
}a[1005];

bool cmp(note x, note y){
	return x.r < y.r;
}

int main(){
	int Case = 1;
	int n, d, x, y, ans = 0, flag = 0;
	double s;
	while(scanf("%d%d",&n,&d) == 2){
		flag=0;
		if(n == 0 && d == 0)    break;
		for(int i = 1; i <= n; i++){
			scanf("%d%d", &x, &y);
			if(y > d || y < (-d)){//不管怎么样雷达都无法覆盖这个点
				flag = 1;//标记一下
			}
			if(flag == 0){
				a[i].l = x - sqrt(1.0*d*d - y*y);//数学原理
				a[i].r = x + sqrt(1.0*d*d - y*y);//同上
			}
		}
		if(flag == 0){
			sort(a+1, a+n+1, cmp);//右端点从小到大排序
			ans = 0;
			s = -(1<<30);
			for(int i = 1; i <= n; i++){
				if(a[i].l > s){//如果某个雷达的左端点>当前右端点，说明这个雷达要选
					ans++;//雷达数+1
					s = a[i].r;//更新当前右端点
				}
			}
		}
		printf("Case %d: ", Case);
		Case++;
		if(flag)
			printf("-1\n");
		else
			printf("%d\n", ans); 
	}
	return 0;
}

PS

理解 + 转换 = 简单