poj1328Radar Installation(贪心—区间选点)

题目链接：

啊哈哈，点我点我

题目：

Radar Installation

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

这道题目是贪心里面的区间选点问题。。贪心策略是：选区间的最右端的点.

思路：首先抽象出这个模型，以海岛为圆心，雷达距离为半径，求出在陆地上的区间，则雷达选在

这个区间之类那么必定能够扫描到这个海岛。。求出所有区间后，就转化成区间选点问题。。

还有就是代码中的那个标准end要用double，我wa了好久。。。

代码为：

#include<cstdio>
#include<cstdio>
#include<algorithm>
#include<cmath>
#define INF 0x3f3f3f3f
using namespace std;

const int maxn=1000+10;

struct Line
{
   double le,ri;
}line[maxn];

bool cmp(Line a,Line b)
{
   if(a.ri!=b.ri)
        return a.ri<b.ri;
   else
        return a.le>b.le;
}

int main()
{
    bool ok;
    int u,v,cas=1;
    int cnt;
	double End;
    int n,d;
    while(~scanf("%d%d",&n,&d))
    {
    	if(n==0&&d==0)   return 0;
        cnt=0;
        ok=false;
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d",&u,&v);
            if(v>d)
                ok=true;
            else
            {
                line[i].le=(double)u-sqrt((double)(d*d-v*v));
                line[i].ri=(double)u+sqrt((double)(d*d-v*v));
            }
        }
        if(ok)
            printf("Case %d: -1\n",cas++);
        else
        {
           sort(line+1,line+1+n,cmp);
           cnt=0;
           End=-INF;
           for(int i=1;i<=n;i++)
           {
               if(line[i].le>End)
               {
                   cnt++;
                   End=line[i].ri;
               }
           }
           printf("Case %d: %d\n",cas++,cnt);
        }
    }
    return 0;
}

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