POJ1328贪心算法

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 75702 Accepted: 16958

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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1开始自己也没想出来 ，直到看完题解才明白还可以再往回走。#include <iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;struct node{    double zuo;    double you;}a[11234];bool cmp(node a,node b){    return a.zuo<b.zuo;}int main(){     int n;     double d;     int fl=1;     while(cin>>n>>d)     {        int sum=1,flag=1;        if(n==0&&d==0) break;        for(int i=0;i<n;i++)        {            int x,y;            scanf("%d%d",&x,&y);            if(y>d||d<0) flag=0;            a[i].you = x+sqrt(d*d-y*y);            a[i].zuo= x-sqrt(d*d-y*y);        }       sort(a,a+n,cmp);       double tt=a[0].you;       for(int i=1;i<n;i++){        if(a[i].you <= tt){        tt = a[i].you;        }        else if(a[i].zuo > tt){        sum++;        tt= a[i].you;        }        }        if(flag==0) sum=-1;        printf("Case %d: %d\n",fl,sum);        fl++;    }    return 0;}