Poj1328-Radar Installation

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

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

 

 真想说句，“啥也不想说了！！！”，qsort函数排序，人家从零开始，我赋值偏要从‘1’开始，调了半天

 1 #include <cstdio>

 2 #include <iostream>

 3 #include <cmath>

 4 #include <cstdlib>

 5 using namespace std;

 6 const int MAXN = 10000 + 10;

 7 

 8 #define FFF freopen("input.txt", "r", stdin)

 9 

10 struct Node

11 {

12     double x, y;

13 } node[MAXN];

14 

15 int cmp(const void *a, const void *b)

16 {

17     return (*(Node *)a).x > (*(Node *)b).x ? 1 : -1;

18 }

19 

20 int main()

21 {

22     FFF;

23     int n;

24     double r;

25     int cas = 1;

26     while(scanf("%d %lf", &n, &r) != EOF && (n+r))

27     {

28         bool tag = false;

29         for(int i = 0; i < n; ++i)

30         {

31             scanf("%lf %lf", &node[i].x, &node[i].y);

32             if(node[i].y > r)

33             {

34                 tag = true;

35             }

36         }

37         if(tag)

38         {

39             printf("Case %d: ", cas++);

40             printf("-1\n");

41             continue;

42         }

43         ///给他进行排序，然后从一端进行扫描

44         qsort(node, n, sizeof(node[0]), cmp);///qsort人家是从下标为0，进行排序，以后注意！！

45 

46         int sum = 1;

47 

48         printf("Case %d: ", cas++);

49 

50         double left[MAXN],righ[MAXN];    //求出每个小岛与坐标轴的交点

51         for(int i=0 ; i < n; i++)

52         {

53             left[i] = node[i].x - sqrt(r*r - node[i].y * node[i].y);

54             righ[i] = node[i].x + sqrt(r*r - node[i].y * node[i].y);

55         }

56 

57         double temp = righ[0];      //将第一个雷达放置在第一个点与坐标轴的右交点

58         for(int i = 0; i < n-1; ++i)

59         {

60             if(left[i+1] > temp)    //这种情况下，不论雷达怎么放置，两小岛都不能在同一雷达范围内，所以必须放置新的雷达

61             {

62                 temp = righ[i+1];

63                 sum++;

64             }

65             else if(righ[i+1] < temp)//当在这种情况下，必须要将雷达所在的位置更新为新小岛与坐标轴的右边界，才能保证

66             {

67                 temp = righ[i+1];    //雷达能够覆盖到这两个岛屿

68             }                        //在其他情况就，不用放置新的雷达，也不用更新雷达的坐标

69         }

70         printf("%d\n", sum);

71     }

72     return 0;

73 }

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