POJ 1328 Radar Installation

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 56591 Accepted: 12768

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：有n艘船及其坐标，在海岸线位置需要安装雷达，已知雷达信号覆盖范围的半径，求最少需要多少个雷达能覆盖完所有船只。若无解，输出-1。

解题思路：求出每艘船能被覆盖的雷达坐标区间，给区间按左边界排序，区间重叠部分设置一个雷达，依次往后。

解题过程：wa点：注意找枚举雷达坐标的精确值，解题过程发现0.01是精确值。细节注意：最小的右区间边界也算在设置雷达范围内。

代码如下：

#include <cstdio>#include <algorithm>using namespace std;const int maxn=1000005;const int mixn=1e-10; int n;struct A{    double x,y;}a[maxn];struct B{    double l,r;}b[maxn];double cmp(B a,B b){    return a.l<b.l;}double findr(int x){double m=1000000.0;for(int i=x;i<n;i++)if(m>b[i].r)m=b[i].r;return m;}int main(){    int ans,t=0,i,p;    double d,low,high,cur,minr;    while(scanf("%d%lf",&n,&d)!=EOF&&(n||d))    {        t++,p=0,ans=1;minr=1000000.0;        if(d<=0.0)p=1;        for(i=0;i<n;i++)        {            scanf("%lf%lf",&a[i].x,&a[i].y);            if(a[i].y>d||a[i].y<0.0)                p=1;        }        if(p)        {            printf("Case %d: -1\n",t);            continue;        }        for(i=0;i<n;i++)        {            low=a[i].x-d;            high=a[i].x;            for(cur=high;cur>=low;cur-=0.01)//这里若采用二分查找临界条件不好判断            {                if((a[i].x-cur)*(a[i].x-cur)+(a[i].y*a[i].y)>d*d)                    break;            }            b[i].l=cur;            low=a[i].x;            high=a[i].x+d;            for(cur=low;cur<=high;cur+=0.01)            {               if((a[i].x-cur)*(a[i].x-cur)+a[i].y*a[i].y>d*d)                        break;            }            b[i].r=cur;if(minr>b[i].r)minr=b[i].r;//找到最小右边界        }sort(b,b+n,cmp);//给区间左边界排序for(i=0;i<n;i++){if(b[i].l<=minr)continue;else{ans++;minr=findr(i);//设置完一个雷达后找新的最小右边界}}        printf("Case %d: %d\n",t,ans);    }    return 0;}