poj1328--Radar Installation

                                                                                      Radar Installation

                                                                                    Time Limit:1000MS    Memory Limit:10000KB    64bit IO Format:%I64d & %I64u

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

题意：雷达区间覆盖。

算法：区间化+贪心

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
#define MAXN 10000

using namespace std;

struct node{
    double l, r;
}Line[MAXN];

bool cmp(node a, node b)//根据结点左区间从小到大排序
{
 	return a.l<b.l;
}

int main()
{
	int n;
	double d;
	int Numcase=0;
	while(~scanf("%d%lf", &n, &d))//本来用cin输入，竟然TLE了
	{
		double x, y;
		int cnt=1;
	    bool flag=false;//如果岛到x轴的距离大于d，那雷达肯定覆盖不了此岛，直接输出-1；
		if(n==0&&d==0) break;
		for(int i=0; i<n; i++)
		{
			cin>>x>>y;
			if(fabs(y)>d) flag = true;
			Line[i].l = x-sqrt(d*d-y*y);//以(x,y)为圆心，d为半径画圆，交x轴两点
			Line[i].r = x+sqrt(d*d-y*y);//分别用 Line[i].l 和 Line[i].r 表示
		}
		if(flag) { cout<<"Case "<<++Numcase<<": "<<-1<<endl; continue; }
        sort(Line, Line+n, cmp);//排序
		double now;
        now = Line[0].r;
        for(int i=1; i<n; i++)
		{
			if(now > Line[i].r)//在所有右结点中，找到距原点最远的点作为now的初始值，开始贪心。
				now = Line[i].r;
			else if(now <Line[i].l)
			{
				now = Line[i].r;
				cnt++;
			}
		}
		cout<<"Case "<<++Numcase<<": "<<cnt<<endl;
	}
	return 0;
}

WA了好几次。。。。。