南阳题目287-Radar

末蓝、 2022-07-26 04:11 209阅读 0赞

## Radar ##

时间限制： 1000 ms | 内存限制： 65535 KB

难度： 3

描述

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.

![20110505064233_82697.jpg][]

输入

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

输出

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.

样例输入

3 2
    1 2
    -3 1
    2 1
    
    1 2
    0 2
    
    0 0

样例输出

Case 1: 2
    Case 2: 1

来源

这个题目就是要反过来想，以每个岛屿为圆心画圆，与坐标轴的两个交点存起来，然后按后面的坐标升序排列。就转换成了一个区间选点的问题。

#include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    struct node {
    	double b,e;
    }s[100100];
    bool cmp(node x,node y)
    {
    	return x.e<y.e;
    }
    int main()
    {
    	int m,n,a,b,i,j,cot=1,tt;
    	double t;
    	while(scanf("%d%d",&m,&n),m+n)
    	{
    		tt=0;
    		int flag=0;
    		for(i=0;i<m;i++)
    		{
    			scanf("%d%d",&a,&b);
    			if(abs(b)>n)
    			flag=1;
    			s[i].b=a-sqrt(n*n-b*b);
    			s[i].e=a+sqrt(n*n-b*b);
    		}
    		sort(s,s+m,cmp);
    		t=s[0].e;
    		for(i=1;i<m;i++)
    		{
    			if(s[i].b>t)
    			{
    				tt++;
    				t=s[i].e;
    			}
    		}
    		if(flag)
    		printf("Case %d: -1\n",cot++);
    		else
    		printf("Case %d: %d\n",cot++,tt+1);
    	}
    	return 0;
     }

[20110505064233_82697.jpg]: http://acm.nyist.net/JudgeOnline/admin/kind/attached/20110505064233_82697.jpg