POJ 1328 - Radar Installation（贪心）

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 45932 Accepted: 10256

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

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题意：在沿海（x轴）放置雷达，雷达侦测半径为d，求使给出的n个小岛全部被覆盖所需要的最少的雷达数量

先求出使每个小岛可侦测的区间，即以小岛为圆心半径为d的圆与x轴的交点之间的部分，就是对于该小岛来说可放置雷达的范围。若没有交点直接输出-1。

求出每个小岛的区间之后就是标准的贪心了

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>using namespace std;struct radar{    double x,y;    double l,r;}a[1111];bool cmp(radar a,radar b){    return a.r<b.r;}int main(){    int n,d,T=1;    while(~scanf("%d%d",&n,&d))    {        if(n==0&&d==0) break;        printf("Case %d: ",T++);        int ok=0;        for(int i=0;i<n;i++)            scanf("%lf%lf",&a[i].x,&a[i].y);        for(int i=0;i<n;i++)        {            if(a[i].y>d)            {                printf("-1\n");                ok=1;                break;            }            a[i].l=a[i].x-sqrt(d*d-a[i].y*a[i].y);            a[i].r=a[i].x+sqrt(d*d-a[i].y*a[i].y);        }        if(ok) continue;        sort(a,a+n,cmp);        double temp=a[0].r;        int ans=1;        for(int i=1;i<n;i++)        {            if(a[i].l>temp)            {                ans++;                temp=a[i].r;            }        }        printf("%d\n",ans);    }    return 0;}