Radar
时间限制：1000 ms | 内存限制：65535 KB
难度：3
- 描述
- 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.
- 输入
- 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 - 输出
- 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.
- 样例输入
-
3 2 1 2 -3 1 2 1 1 2 0 2 0 0
- 样例输出
-
Case 1: 2 Case 2: 1
View Code
1 2 #include<stdio.h> 3 #include<math.h> 4 #include<string.h> 5 #include<algorithm> 6 using namespace std; 7 struct dy 8 { 9 double left;//**左交点**// 10 double right;//**右交点**// 11 }w[1001]; 12 bool comp(dy a,dy b)//**按左交点坐标从小到大排序**// 13 { 14 if(a.left<b.left) return true; 15 return false; 16 } 17 int main() 18 { 19 int n,r,x,y,i,count,num=1; 20 double len,t; 21 while(~scanf("%d %d",&n,&r)&&(n,r)) 22 { 23 memset(w,0,sizeof(w)); 24 count=1;//**从第一个点开始，所以计数器初值为1**// 25 for(i=0;i<=n-1;i++) 26 { 27 scanf("%d %d",&x,&y); 28 len=sqrt(((double)r*r-(double)y*y));//**结合图形，勾股定理**// 29 w[i].left=(double)x-len;//**左交点的坐标**// 30 w[i].right=(double)x+len;//**右交点的坐标**// 31 } 32 for(i=0;i<=n-1;i++) 33 { 34 if(y>r)//**如果不能完全覆盖**// 35 { 36 printf("Case %d: -1\n",num++); 37 break; 38 } 39 } 40 sort(w,w+n,comp); 41 t=w[0].right; 42 for(i=1;i<=n-1;i++) 43 { 44 if(w[i].left>t)//**如果后一个点的左交点大于前一个点的右坐标，说明两点没有公共区域**// 45 { 46 count++; 47 t=w[i].right; 48 } 49 else 50 { 51 if(w[i].right<t)//**如果后一个点的右交点小于前一个点的右坐标，说明后一个点的覆盖区域被前一个点包含了**// 52 { 53 t=w[i].right;//**保证后一个点被覆盖**// 54 } 55 } 56 } 57 printf("Case %d: %d\n",num++,count); 58 } 59 return 0; 60 }
