OpenJ_Bailian 1328

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Radar Installation

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice OpenJ_Bailian 1328

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

题意：

       在一个坐标平面上有n个点，他们的y坐标都大于0。雷达可以覆盖范围为d的区域，问在y=0这条直线上至少设置多少个雷达，可以覆盖完所有给出的n个点。

输入：

       多组数据，第一行是n和d，之后n行是每个点的x坐标与y坐标。

输出：

       最少的雷达数。

分析：

       首先进行转化，求出每一个点所对应的一段y=0上的区间，在这段区间上放置雷达将会覆盖这个点。然后再对这n个区间按照左端的从小到大进行排序。r_index变量表示当前所有区间的最右边界，如果下一个区间的左端点不比r_index大，则用该区间的有端点的值更新r_index = min(r_index,右端点的值)。如果下一个区间的左端点比r_index大，则ans++，并且它的右端点的值赋给r_index。按这样的方式遍历完所有区间就可以得出答案。

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <vector>
 5 #include <string>
 6 #include <algorithm>
 7 #include <cmath>
 8 using namespace std;
 9 const int MAX_N = 1000;
10 struct scope{double l,r;}s[MAX_N + 10];
11 bool cmp(const scope& s1,const scope& s2){
12     return s1.l < s2.l;
13 }
14 int main(){
15     int n,R;
16     int Case = 0;
17     while(scanf("%d%d",&n,&R),n + R){
18         int ans = 1;
19         int x,y;
20         for(int i = 0 ; i < n ; i++){
21             scanf("%d%d",&x,&y);
22             s[i].l = x - sqrt(R * R - y * y);
23             s[i].r = x + sqrt(R * R - y * y);
24             if(y > R || R <= 0) ans=-1;
25         }
26         sort(s,s + n,cmp);
27         double r_index = s[0].r;
28         for(int i = 1 ; i < n && ans != -1 ; i++){
29             if(s[i].l > r_index){
30                 ans++;
31                 r_index = s[i].r;
32             }
33             else if(s[i].r < r_index){
34                 r_index = s[i].r;
35             }
36         }
37         printf("Case %d: %d\n",++Case, ans);
38     }
39     return 0;
40 }

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OpenJ_Bailian 1328

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原文地址：http://www.cnblogs.com/cyb123456/p/5783288.html