POJ 1328 Radar Installation （贪心）

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

题意

海上有n多岛，在海岸线上（x轴）建一个雷达能覆盖到与它距离不超过d的岛，求覆盖所有岛的最小雷达数。

思路

以岛屿为圆心，d为半径画圆，交x轴于s,e两点，然后以线段末尾为排序标准对所有这样的线段排序，从左至右依次判断，假设雷达放置在一个线段的末端，然后忽略它所能涉及到的岛屿。

AC 代码

#include <iostream>#include<stdio.h>#include<math.h>#include<string.h>#include<algorithm>#include<stdlib.h>#include<queue>using namespace std;struct point{    double x,y;    double s,e;    bool operator<(const point &o)const    {        return e<o.e;    }} land[1005];int solve(int n,int d){    for(int i=0; i<n; i++)  //计算所有的线段    {        double t=sqrt(d-land[i].y*land[i].y);        land[i].s=land[i].x-t;        land[i].e=land[i].x+t;    }       sort(land,land+n);      //排序    int ans=1;    double r=land[0].e;    for(int i=1; i<n; i++)    {        if(land[i].s>r)        {            ++ans;            r=land[i].e;        }    }    return ans;}int main(){    int n,d,cat=1;    while(~scanf("%d%d",&n,&d)&&(n||d))    {        bool flag=false;        for(int i=0; i<n; i++)        {            scanf("%lf%lf",&land[i].x,&land[i].y);            if(land[i].y>d)flag=true;   //是否可以检测到该岛屿        }        if(flag)printf("Case %d: -1\n", cat++);        else printf("Case %d: %d\n", cat++, solve(n,d*d));    }    return 0;}