POJ1328 Radar Installation

Posted 2020-10-02 嘒彼小星

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 89661		Accepted: 20143

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

Source

Beijing 2002

 

【题解】

每个点对应一段放置的区间，于是问题转化成了区间点覆盖问题。贪心即可

 

 1 #include <iostream>
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cstdlib>
 5 #include <string>
 6 #include <algorithm>
 7 #include <cmath>
 8 #define min(a, b) ((a) < (b) ? (a) : (b))
 9 
10 const int INF = 0x7fffffff;
11 const int MAXN = 1000 + 10;
12 
13 inline void read(int &x)
14 {
15     x = 0;char ch = getchar(),c = ch;
16     while(ch < ‘0‘ || ch > ‘9‘)c = ch, ch = getchar();
17     while(ch <= ‘9‘ && ch >= ‘0‘)x = x * 10 + ch - ‘0‘, ch = getchar();
18     if(c == ‘-‘)x = -x;
19 }
20 
21 int n, R, cnt[MAXN], ans, t, ok;
22 double l[MAXN], r[MAXN], now;
23 
24 bool cmp(int a, int b)
25 {
26     return l[a] < l[b];
27 }
28 
29 int main()
30 {
31     while(scanf("%d%d", &n, &R) && (n + R))
32     {
33         ans = now = ok = 0;
34         register int x,y;
35         for(register int i = 1;i <= n;++i)
36         {
37             cnt[i] = i;
38             read(x);read(y);
39             if(R < y) ans = -1,ok = 1;
40             double tmp = sqrt(R * R - y * y);
41             l[i] = x - tmp;
42             r[i] = x + tmp;
43         }
44         if(ok)goto L1;
45         std::sort(cnt + 1, cnt + 1 + n, cmp);
46         now = -INF;
47         for(register int i = 1;i <= n;++ i)
48             if(l[cnt[i]] > now) ans ++, now = r[cnt[i]];
49             else now = min(now, r[cnt[i]]);
50 L1:        ;
51         if(!ans) ans = -1;
52         printf("Case %d: %d\n", ++t, ans);
53     } 
54     return 0;
55 } 

POJ1328