【POJ 1328】Radar Installation

Radar Installation

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

题意：雷达只能放在x轴上，给出雷达半径，和岛屿坐标，问至少按多少个雷达可以覆盖全岛：

解题思路：根据岛屿到x轴的距离和半径r求出在x上的一个区间，区间内雷达都能扫到该岛公式：
左区间=x-sqrt(r*r-y*y):
右区间=x+sqrt(r*r-y*y):
按照右区间从小到大排序
要考虑 y>d 和d<0的情况
详细看AC代码 ：

#include<stdio.h>#include<math.h>#include<algorithm>using namespace std;struct node{    double a,b;}st[1000+11];bool cmp(node i,node j){    return i.a<j.a;}int main(){    int n,i,k=0;    double x,y,d;    bool flag;    while(scanf("%d%lf",&n,&d)!=EOF&&(n||d))    {        k++;        flag=false;        for(i=0;i<n;i++)        {        scanf("%lf%lf",&x,&y);        if(y>d||d<0)         flag=true;        st[i].a=x-sqrt(d*d-y*y);        st[i].b=x+sqrt(d*d-y*y);        }         if(flag)        {            printf("Case %d: -1\n",k);            continue;        }        sort(st,st+n,cmp);        double max=st[0].b;int sum=1;;        for(i=1;i<n;i++)        {        if(st[i].b<=max)         {        max=st[i].b;        }        else if(st[i].a>max)            {                sum++;                max=st[i].b;            }        }        printf("Case %d: %d\n",k,sum);    }    return 0;}