区间选点问题（贪心）

贪心。

一开始的思路：

图中ABCD为海岛的位置。假设本题半径为2（符合坐标系），那么A点坐标为（1，1）以此类推。

在题中，记录每个点的坐标，并把一个点新加一个标记变量以标记是否被访问过。

首先以A为圆心，r为半径做圆，交x轴与E1（右）点，做出如图中绿色虚线圆。然后以E1为圆心，半径为r做圆。看此时下一点（B）是否在该圆中。如果在，那么将该点标记变量变为true，再判下一点（C），如果不在那么就新增一个雷达。

还是原来的贪心思路，仍然先排序，但排序的准则是按红色圆与x轴左交点的先后顺序。

如图，如果B点圆（且如此称呼）的左交点（J1点）在A点圆右交点（E1）左侧，那么B点一定涵盖在A点圆内部。再如图，如果D点圆的左交点（图中未标示）在A点圆的右交点（未标示）右，则D点不在A点圆中。

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 38208		Accepted: 8483

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

#include
#include
using namespace std;
typedef struct
{
	double l,r;
}in;
int cmp(const void *a, const void *b)
{
	return (*(in *)a).l >= (*(in *)b).l ? 1:-1;
}
int main()
{
	int n,d,i,x,y,sw,re,count = 1;
	double pre;
	in p[1000];
	while(1)
	{
		cin>>n>>d;
		if(n == 0 && d==0) break;
		sw = 1;
		for(i=0;i>x>>y;
			if(d>=y&&sw==1)
			{
				p[i].l = x-sqrt((double)d*d - (double)y*y);
				p[i].r = x+sqrt((double)d*d - (double)y*y);
			}
			else
			{
				sw = 0;
			}
		}
		if(sw == 0)
		{
			cout<<"Case "<pre)
			{
				re++;
				pre = p[i].r;
			}
			else
			{
				if(p[i].r

Case 2: 1

Source

Beijing 2002

题目的意思就是给你一个坐标轴，雷达在x轴上，岛屿分布在x轴上方，给你岛屿的坐标以及雷达的最大扫描面积，求最少用几个雷达可以将所有的岛屿覆盖！

思路：

以岛为圆心，以d为半径画圆（d是雷达的辐射半径），其与x轴相交的区间为一个区  这样就变成了在区间内找最少的点问题了