[2016-02-04][POJ][1328][Radar Installation]

[2016-02-04][POJ][1328][ Radar Installation ]

Radar Installation

Time Limit: 1000MS

Memory Limit: 10000KB

64bit IO Format: %I64d & %I64u

Submit Status

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

• 时间:2016-02-04 22:01:05 星期四
  
• 题目编号:POJ 1328
  
• 题目大意:
• 在坐标系上给定若干个点,在x轴上设置雷达,使得每个点都被雷达扫描,
• 求最少雷达数目
• 分析:
• 可以计算出每一个点,对应的雷达在x轴的放置范围,
• 那么题目就转化为,给定若干个区间,求最少点的数目,使得每一个区间都有一个点
• 即,相交区间的数目
• 方法:
• 排序 按右端点 小→大(相等时,按左端点大→小)对区间进行排序,
  
• 记录上一个区间,
• 选择没有和上一个区间相交的区间
• 计数+1
• 重复

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

#include<cstdio> #include<algorithm> #include<cmath> using namespace std; const int maxn = 1000 + 10; struct room{ double s,e; bool operator < ( const room a) const { return e < a.e || (e == a.e && s > a.s); } }r[maxn]; int main(){ int n,d; int cnt = 0; //freopen("out.txt","w",stdout); while (~ scanf ( "%d%d" ,&n,&d) && (n || d)){ int x,y; int res = 1; for ( int i = 0; i < n ;i++ ){ scanf ( "%d%d" ,&x,&y); r[i].s = x - sqrt ((( double )d*d - y*y)); r[i].e = x + sqrt ((( double )d*d - y*y)); if (y > d) res = -1; } if (~res) { sort(r,r+n); room pre = r[0]; for ( int i = 1; i < n;i++){ if (r[i].s > pre.e){ res++; pre = r[i]; } } } printf ( "Case %d: %d\n" , ++cnt, res); } return 0; }

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