POJ 1328：Radar Installation

标签：poj

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

Output

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：该题题意是为了求出能够覆盖所有岛屿的最小雷达数目，每个小岛对应x轴上的一个区间，在这个区间内的任何一个点放置雷达，则可以覆盖该小岛，区间范围的计算用[x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)];这样，问题即转化为已知一定数量的区间，求最小数量的点，使得每个区间内斗至少存在一个点。每次迭代对于第一个区间， 选择最右边一个点， 因为它可以让较多区间得到满足， 如果不选择第一个区间最右一个点（选择前面的点）， 那么把它换成最右的点之后， 以前得到满足的区间， 现在仍然得到满足， 所以第一个区间的最右一个点为贪婪选择， 选择该点之后， 将得到满足的区间删掉， 进行下一步迭代， 直到结束。

以左区间为准：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>

using namespace std;

const int M = 1000 + 5;
double x, y;
double dis;           //圆的半径
double temp;
int ans;             //用于统计结果
int n;               //点的个数
int flag;            //用于判断

struct node
{
    double left;       //左区间
    double right;     //右区间 
} island[M];

bool cmp(node a, node b)   //按左区间从小到大排序
{
    return a.left<b.left;
}

int main()
{
    int cas=0;
    while(scanf("%d%lf", &n, &dis) && n && dis)
    {
        flag=0;
        ans=0;
        cas++;
        for(int i=0; i<n; i++)
        {
            scanf("%lf%lf", &x, &y);
            if(y>dis||y<0)               //不符合的点
                flag=1;
            island[i].right = x+sqrt(dis*dis-y*y);      
            island[i].left = x-sqrt(dis*dis-y*y);
        }
        if(flag==1)
            printf("Case %d: -1\n",cas);
        else
        {
            sort(island, island+n, cmp);   //sort排序
            temp = island[0].right;       
            ans=1;
            for(int i=1; i<n; i++)
            {
                if(island[i].right<temp)          //说明这个区间被上个区间所包含，只要小区间满足的点，大区间肯定也满足
                    temp=island[i].right;
                else if(island[i].left>temp)     //说明这两个区间没有重合部分，也就不能用一个圆表示，则结果加一
                {
                    ans++;
                    temp=island[i].right;
                }
            }
            printf("Case %d: %d\n", cas, ans);
        }
    }

    return 0;
}

一右区间为准：（这是我同学写的右区间）

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<algorithm>

using namespace std;

struct node
{
	double a,b;
}s[1005],l[1005];

int cmp(node x,node y)
{
	return x.b<y.b;
}

int main()
{
	int n,m,k=0;
	while(cin>>n>>m)
	{
		if(n==0&&m==0)   break;
		int i,j,p=0;
		for(i=0;i<n;i++)
		{
			scanf("%lf%lf",&s[i].a,&s[i].b);
			if(s[i].b<0||s[i].b>m)
				p=1;
			l[i].a=s[i].a-sqrt(m*m-s[i].b*s[i].b);
			l[i].b=s[i].a+sqrt(m*m-s[i].b*s[i].b);
		}
		sort(l,l+n,cmp);
		int sum=1; double r=l[0].b;
		for(i=1;i<n;i++)
		{
			if(l[i].a>r)
			{
				r=l[i].b;
				sum++;
			}
		}

		printf("Case %d: ",++k);
		if(p)
			cout<<-1<<endl;
		else
			cout<<sum<<endl;
	}
	return 0;
}

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POJ 1328：Radar Installation

标签：poj

原文地址：http://blog.csdn.net/u013487051/article/details/37569807