Radar Installation

Radar Installation
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%lld & %llu
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Status

Practice

POJ 1328
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
计算出每个岛雷达所能覆盖的半径然后按照右区间或左区间排序

#include <cmath>#include <algorithm>#include <cstdio>using namespace std;struct node{    double left,right;}a[1001];bool cmp(node a,node b){    return a.left < b.left;}int main(){    double x,y,d,temp;    int i,cnt,n,k=0;    bool flag;    while(scanf("%d%lf",&n,&d)&&!(n==0&&d==0))    {        k++;        flag=false;        for(i=0;i<n;i++){            scanf("%lf%lf",&x,&y);            if(y > d||d<0){                flag = true;            }            a[i].right = x+sqrt(d*d-y*y);            a[i].left = x-sqrt(d*d-y*y);        }        if(flag){        printf("Case %d: -1\n",k);            continue;        }        sort(a,a+n,cmp);        temp=a[0].right;        cnt=1;            for(i=1;i<n;i++){            if(a[i].right <= temp){//这里比较难理解就是说雷达要尽可能向右放但是不可以有漏掉的岛            temp = a[i].right;        }           else if(a[i].left > temp){            cnt++;            temp = a[i].right;            }            }            printf("Case %d: %d\n",k,cnt);        }        return 0;}