poj 1328 贪心

http://poj.org/problem?id=1328

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 90991		Accepted: 20407

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

   没记错得话也是某届HNACM省赛的原题- -当时以为是计算几何什么的

   一个很容易猜到的贪心是从左至右遍历，遇见没覆盖的点时尽可能的在保证覆盖此点的情况下往右边建。但是这是一个错误的贪心方案，圆心向右移动的同时最高点也将向右移动，这有可能导致原本可以覆盖住的点漏了出来导致答案错误，很忧桑。正解是对于每一个点考虑圆心可能的位置，显然每个点对应一条在x轴上的线段，这就转化为了一个经典的贪心问题，用最小点覆盖所有区间，按r排序后贪心建在最右边即可。

  

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
#define eps 1e-8
struct node{double x,y;}P[1005];
bool cmp(node A,node B)
{
    if(A.y==B.y) return A.x<B.x;
    return A.y<B.y;
}
int main()
{
    int N,D,i,j,k=0;
    while(cin>>N>>D&&(N||D)){int ans=0,ok=0;
    double x,y;
        for(i=0;i<N;++i)
        {
            scanf("%lf%lf",&x,&y);
            double r=sqrt(D*D-y*y);
            P[i].x=x-r;
            P[i].y=x+r;
            if(abs(y)>D)ok=1;
        }
        sort(P,P+N,cmp);
        double en=-999999999;
        for(i=0;i<N;++i)
        {
          if(P[i].x>en){
            ans++;
            en=P[i].y;
          }
        }
        if(ok) ans=-1;
        printf("Case %d: %d
",++k,ans);
    }
    return 0;
}