POJ 1328Radar Installation(贪心好标题)

POJ 1328Radar Installation(贪心好题目)
Radar InstallationTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 44740 Accepted: 9922

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

题目大意：平面上有一些点，现要求用一些圆心在x轴上的圆（雷达）来覆盖这些点，问最少需要多少雷达
解题思路：每个雷达尽量覆盖更多的点，先对点按x坐标排序，在最左边建一个雷达，在雷达左边的点可以覆盖，只需要将雷达左移即可。但是在雷达右边的却需要新建立雷达。具体实现见代码。
题目地址：Radar Installation
AC代码：

#include<iostream>#include<cstdio>#include<cmath>#include<algorithm>using namespace std;const double eps=1e-6;struct mq{    double x;    double y;};mq node[1002];bool cmp(mq a,mq b){    if(a.x<b.x) return true;    return false;}double dis(double xx,double x,double y){    return sqrt(y*y+(xx-x)*(xx-x));}int main(){    int tes=0,n,i;    double d;    double xx;    while(scanf("%d%lf",&n,&d))    {        int res=1,flag=0;        if(n==0&&d<eps) break;        for(i=0;i<n;i++)        {            scanf("%lf%lf",&node[i].x,&node[i].y);            if(node[i].y-d>eps) flag=1;        }        printf("Case %d: ",++tes);        if(flag) {puts("-1"); continue;}  //说明圆包括不了        sort(node,node+n,cmp);        xx=node[0].x+sqrt(d*d-node[0].y*node[0].y);        for(i=1;i<n;i++)        {            double tmp;            tmp=node[i].x+sqrt(d*d-node[i].y*node[i].y);            if(dis(xx,node[i].x,node[i].y)>d)  //说明覆盖不了            {                if(tmp>xx) res++;  //需要重新建立一个雷达                xx=tmp;               }        }        printf("%d\n",res);    }    return 0;}/*2 5-3 4-6 3*///47MS

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