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Poj1328-Radar Installation

时间:2014-11-03 01:16:29      阅读:294      评论:0      收藏:0      [点我收藏+]

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. 
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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
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 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);
45 
46         int sum = 1;
47 
48         printf("Case %d: ", cas++);
49 
50         double left[MAXN],righ[MAXN];
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];
58         for(int i = 0; i < n-1; ++i)
59         {
60             if(left[i+1] > temp)
61             {
62                 temp = righ[i+1];
63                 sum++;
64             }
65             else if(righ[i+1] < temp)
66             {
67                 temp = righ[i+1];
68             }
69         }
70         printf("%d\n", sum);
71     }
72     return 0;
73 }
View Code

先睡觉吧,明天再说!!!

Poj1328-Radar Installation

原文:http://www.cnblogs.com/ya-cpp/p/4070342.html

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