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
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.
![](http://poj.org/images/1328_1.jpg)
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
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
Source
大意:河边有很多岛,要在沿岸建雷达,将点化成在沿岸的坐标,进行贪心,注意当排序完左之后,要考虑如果有点的右在ax的右,那么ax就变成该店,因为这个,一直没A~~~
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; const int maxn = 1000; struct X{ double left; double right; }dp[maxn]; int x[maxn],y[maxn]; bool cmp(X i,X j){ return i.left<j.left;} int main(){ int n,d,i, flag1 = 1; while(~scanf("%d%d",&n,&d)&&n&&d){ int flag2 = 0; for( i = 1; i <= n ; i++){ scanf("%d%d",&x[i],&y[i]); if(y[i] > d||d<0||y[i]<0) flag2 = 1; dp[i].left = x[i] - sqrt((double)(d*d-y[i]*y[i])); dp[i].right = x[i] + sqrt((double)(d*d-y[i]*y[i])); } if(flag2 == 1) { printf("Case %d: -1 ",flag1); flag1++; continue;} sort(dp+1,dp+n+1,cmp); int ax = 1,flag = 1; for(i = ax + 1; i <= n ; i++){ if(dp[i].right < dp[ax].right){ ax = i; } else if(dp[i].left > dp[ax].right){ ax = i; flag++; } } printf("Case %d: %d ",flag1,flag); flag1++; } return 0; }