| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 60277 | Accepted: 13583 |
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.
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
使用贪心算法解决。
1 #include <iostream> 2 #include<math.h> 3 #include <algorithm> 4 #include <stdlib.h> 5 using namespace std; 6 struct point 7 { 8 double left, right; 9 }points[1000], temp; 10 11 bool operator < (point a, point b) 12 { 13 return a.left < b.left; 14 } 15 int main() { 16 int num = 0; 17 double rad = 0; 18 cin >> num >> rad; 19 int casenum=1; 20 while (num || rad) { 21 22 23 int rnum=1; 24 int wrong = 0; 25 26 for (int i = 0; i < num; i++) { 27 double x=0; 28 double y=0; 29 cin >> x >> y; 30 31 if (y > rad) { 32 //岛屿纵坐标大于雷达半径 33 wrong = 1; 34 }else{ 35 //以岛屿为圆心，做圆，与x轴的交点 36 points[i].left=x-sqrt((rad*rad-y*y)*1.0); 37 points[i].right=x+sqrt((rad*rad-y*y)*1.0); 38 39 } 40 } 41 if (wrong) { 42 //雷达不能覆盖 43 cout<<"Case "<<casenum++<<": "<<-1<<endl; 44 } else { 45 //按照岛屿画圆与x轴的左交点排序 46 sort(points, points + num); 47 temp=points[0]; 48 for(int i=1;i<num;i++){ 49 // 50 if(points[i].left>temp.right)//temp.right为公共区间的右端点，若下一个区间的左端点大于temp.right则该区间与以上的公共区间没有公共部分 51 { 52 rnum++;//雷达个数加1 53 temp=points[i];//更新公共区间右端点 54 }else if(points[i].right<temp.right){ 55 temp=points[i];//更新公共区间右端点 56 } 57 } 58 59 cout<<"Case "<<casenum++<<": "<<rnum<<endl; 60 } 61 62 cin >> num >> rad; 63 } 64 return 0; 65 }