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
给定海岛个数、雷达半径以及各海岛坐标，求能覆盖所有海岛的最小雷达数。
贪心策略依然是从左往右，尽量让每颗雷达覆盖最大岛屿数。
对于每个点先求出以该点为圆心，d为半径的圆在x轴上的交点，左右交点就是覆盖此海岛的雷达所在的区间。对于每个区间，按照区间的右端点排序。
从最左边开始贪心，对于第一个右端点x0，所有左端点<x0的海岛都会与x0共用一个雷达
#include<iostream> #include<cmath> #include<cstring> #include<cstdio> #include<algorithm> #define inf 0x3f3f3f3f #define ll long long using namespace std; struct node { int x,y; }q[1010]; struct xnode { double l,r; }xn[1010]; bool cmp(xnode a,xnode b) { if(a.r==b.r) return a.l<b.l; return a.r<b.r; } bool visited[1010]; int main() { int n,d; int kcase=1; while(cin>>n>>d) { if(n==0) return 0; bool flag=0; for(int i=1;i<=n;i++) { cin>>q[i].x>>q[i].y; if(q[i].y>d) flag=1; } if(flag==1||d==0) { cout<<"Case "<<kcase++<<": "<<-1<<endl; continue; } for(int i=1;i<=n;i++) { double len=sqrt(1.0*d*d-q[i].y*q[i].y); xn[i].l=q[i].x-len; xn[i].r=q[i].x+len; } sort(xn+1,xn+n+1,cmp); int ans=0; memset(visited,0,sizeof(visited)); for(int i=1;i<=n;i++) if(!visited[i]) { visited[i]=1; ans++; for(int j=i+1;j<=n;j++) if(!visited[j]&&xn[j].l<=xn[i].r) visited[j]=1; } cout<<"Case "<<kcase++<<": "<<ans<<endl; } return 0; }