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.
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
该题题意是为了求出能够覆盖所有岛屿的最小雷达数目
这题一开始根本没有贪心思路，想不到
qu[i].left=x-sqrt(r*r-y*y);
qu[i].right=x+sqrt(r*r-y*y);
转化为一个区间问题。
关于这个的合理性 你们画个图就好了 ，表示不会电脑画图
#include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<queue> #include<cctype> using namespace std; struct node { double left,right; }qu[1010]; int cmp(node a,node b) { return a.right<b.right; } int main() { int n,k=1; double r; while(scanf("%d%lf",&n,&r)!=EOF){ if (n==0 && r==0 ) break; int flag=1; double x,y; for (int i=0 ;i<n ;i++){ scanf("%lf%lf",&x,&y); if (!flag) continue; if (y>r) { flag=0; continue; } qu[i].left=x-sqrt(r*r-y*y); qu[i].right=x+sqrt(r*r-y*y); } sort(qu,qu+n,cmp); printf("Case %d: ",k++); if (!flag) { printf("-1\n"); continue; } int sum=0; double temp=-10000000; for (int i=0 ; i<n ;i++){ if (qu[i].left>temp){ sum++; temp=qu[i].right; } } printf("%d\n",sum); } return 0; }
