poj1328Radar Installation 贪心

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 64472		Accepted: 14497

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

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

using namespace std;

struct node
{
    double L,R;
} p[1005];
int cmp(node p1,node p2)
{
    return p1.L<p2.L;
}
int main()
{
    int n,d,num=0;
    while(cin>>n>>d)
    {
        num++;
        if(n==0&&d==0)
            break;
        int flag=1;
        for(int i=0; i<n; i++)
        {
            int u,v;
            cin>>u>>v;
            if(flag==0)
	    continue;
            if(d<v)      //注意半径可以取负的,所以不能用d*d<v*v比较
            {
                flag=0;
            }
            else
            {
                p[i].L=(double)u-sqrt((double)(d*d-v*v));
                p[i].R=(double)u+sqrt((double)(d*d-v*v));
            }
        }
        if(flag==0)
        {
            printf("Case %d: -1\n",num);
            continue;
        }

        sort(p,p+n,cmp);
        double x=p[0].R;
        int sum=1;
        for(int i=1; i<n; i++)
        {
			if(p[i].R<x)
            {
                x=p[i].R;
            }
            else if(x<p[i].L)
            {
                sum++;
                x=p[i].R;
            }
        }
        printf("Case %d: %d\n",num,sum);
    }
}

/*把每个岛屿来当做雷达的圆心，半径为d，做圆，与x轴会产生两个焦点L和R，这就是一个区间；
首先就是要把所有的区间找出来，然后x轴从左往右按L排序，再然后就是所谓的贪心
把那些互相重叠的区间去掉就行了区间也就是雷达；*/

/*
 3 -3
 1 2
 -3 2
 2 1
Case  ... -1;
 */
//按R进行从左到右排序
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>

using namespace std;

struct node
{
    double L,R;
} p[1001];
int cmp(node p1,node p2)
{
    return p1.R<p2.R;
}
int main()
{
    int n,d,num=0;
    while(cin>>n>>d)
    {
        num++;
        if(n==0&&d==0)
            break;
        int flag=0;
        for(int i=0; i<n; i++)
        {
            int u,v;
            cin>>u>>v;
            if(d<v)
            {
                flag=1;

            }
            else if(flag==0)
            {
                p[i].L=u-sqrt(d*d-v*v);
                p[i].R=sqrt(d*d-v*v)+u;
            }
        }
        if(flag)
        {
            printf("Case %d: -1\n",num);
            continue;
        }

        sort(p,p+n,cmp);
        double  xR=p[0].R;
        double  xL=p[0].L;
        int sum=1;
        for(int i=1; i<n; i++)
        {
            if(p[i].L<=xR)
            {
            }
            else if(p[i].L>xR)
            {
                xR=p[i].R;
                sum++;
            }
        }
        printf("Case %d: %d\n",num,sum);
    }
}