poj1328Radar Installation(貪心—區間選點)

poj1328Radar Installation(貪心—區間選點)

題目鏈接：

啊哈哈，點我點我

題目：

Radar Installation Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 52037 Accepted: 11682

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

這道題目是貪心裡面的區間選點問題。。貪心策略是：選區間的最右端的點.

思路：首先抽象出這個模型，以海島為圓心，雷達距離為半徑，求出在陸地上的區間，則雷達選在

這個區間之類那麼必定能夠掃描到這個海島。。求出所有區間後，就轉化成區間選點問題。。

還有就是代碼中的那個標准end要用double，我wa了好久。。。

代碼為：

#include
#include
#include
#include
#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.rib.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;
}