poj 1328 Radar Installation（贪心，线段重叠）

poj 1328 Radar Installation

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

 思路 ：

             1. 将每个小岛看做中心，雷达就变成了范围，就转成线段重叠问题，我们然尽量多的线段重叠，就可以安装最少的雷达了。

             2.还有一些注意事项会在代码提醒。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include <cmath>
using namespace std;

const double dat = 0.00000001;

typedef struct node{
double l,r;
}node;

node a[1010];

int cmp(node x,node y )
{
return x.l < y.l; // 按照左端点排序
}
int main()
{
int n;
int index = 0;//样例下标
double m;
while(scanf("%d%lf",&n,&m)&& m+n)
{
bool flag = 1;
for(int i = 0; i < n; i++)
{
double x,y;
cin >> x >> y;
if(y > m)
{
flag = 0;
continue;//不能直接跳出，得让样例输完。
}
a[i].l = x - sqrt(m*m - y*y);
a[i].r = x + sqrt(m*m - y*y);
}
if(flag == 0)
{
printf("Case %d: -1\n",++index);
continue;
}
sort (a,a+n,cmp);
int num = 1;
double tmpr = a[0].r;//记录最远合适的 r 为 tmpr
for(int i = 1; i < n; i++)
{
if(a[i].r < tmpr)//此时仍然重叠 改变tmpr 为重叠部分最小值
{
tmpr = a[i].r;
}
else if(tmpr < a[i].l)//不再重叠 必须加 雷达了
{
tmpr = a[i].r;
num ++;
}
//其他情况就不用变了。

4000
}
printf("Case %d: %d\n",++index,num);
}
return 0;
}