poj 1328 Radar Installation (简单的贪心)

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 42925		Accepted: 9485

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

Source

Beijing 2002

题目链接：http://poj.org/problem?id=1328

题意：有一条直的海岸线，上面有雷达。以海岸线为x轴，x轴上面为海，下面为岸。海里面有很多岛屿。已知雷达的观测半径。问最少建多少个雷达能把所有岛屿都覆盖到雷达

           的侦测范围内。如果不能覆盖全部的岛屿，按样例输出Case数和-1，反之输出Case数和最少需要建立的雷达数。

分析：首先算出经过每个雷达且平行于x轴的弦：假设雷达覆盖半径为d。则弦的左端点为x-sqrt (d*d-y*y),右端点为x+sqrt(d*d-y*y)。然后按左端点从小到大排序。由于不能受右

           端点的影响，将弦的左右端点横坐标用结构体存起来。然后，将第一个雷达放在排序后的弦投影在x轴的右端点temp。从过第2个岛屿的弦的投影开始扫描，如果右端点

           <temp即右端点在雷达左边。把雷达移动到此右端点，此时就能使雷达既覆盖到这个岛屿，又覆盖到前面的岛屿。如果左端点在雷达右边，则不能覆盖，需要再建立一个

            雷达。然后把新雷达建在此时弦右端点投影到x轴的坐标。忽略前面的，对后面的岛屿进行同样的操作。（贪心思想）

注意：半径和坐标都有可能是小数，所以用double类型比较好，不然用int又要注意计算时*1.000变为浮点数。

代码：

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

struct point
{
    double left;
    double right;
}p[5012];
bool flag;
int cnt;

bool cmp(point &x,point &y)
{
    return x.left<y.left;
}

int main()
{
    int cases=0,i,n;
    double d;
    while(scanf("%d%lf",&n,&d)&&n&&d)
    {
        flag=true;
        double a,b;
        for(i=0;i<n;i++)
        {
            scanf("%lf%lf",&a,&b);
            if(fabs(b)>d)
	    flag=false;
            p[i].left=a-sqrt(d*d-b*b);
            p[i].right=a+sqrt(d*d-b*b);
        }
	printf("Case %d: ",++cases);
	if(!flag)
	    puts("-1");
	else
	{
	    sort(p,p+n,cmp);
	    double temp=p[0].right;
	    cnt=1;
	    for(i=1;i<n;i++)
	    {
		if(p[i].right<temp)
		   temp=p[i].right;
		else if(p[i].left>temp)
	        {
		   cnt++;
		   temp=p[i].right;
		}
	    } 
	    printf("%d
",cnt);
	}
    }
    return 0;
}


//AC

 

 

11920732

fukan

1328

Accepted

188K

32MS

C++

981B

2013-08-05 00:20:15