1328:Radar Installation（贪心算法）

总时间限制:: 1000ms
内存限制:: 65536kB

描述

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

输入

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

输出

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.

样例输入

样例输出

Case 1: 2
Case 2: 1

先给贴上测试数据，推荐参考。大佬给的。

2 5
-3 4
-6 3


4 5
-5 3
-3 5
2 3
3 3

20 8
-20 7
-18 6
-5 8
-21 8
-15 7
-17 5
-1 5
-2 3
-9 6
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 7
9 6
10 5
0 0

2 3
0 2
2 3

2 3
0 2
1 3

3 3
1 2
-3 2
2 4

8 5
2 4
-4 4
-3 3
-3 1
-3 0
-1 0
0 5
6 0

3 0
1 2
-3 1
2 1

3 2
1 2
-3 1
2 1

1 2
0 2


2 3
0 2
2 3

4 -5
4 3
4 3
2 3
6 -9



3 -3
1 2
-3 2
2 1

6 2
1 2
1 2
1 2
-3 1
2 1
0 0

1 2
0 2

2 3
0 2
1 3

3 10
1 10
2 3
4 5

3 5
1 10
2 3
4 5

4 7
1 10
2 3
4 5
0 0

3 9
1 10
2 3
4 5
0 0

================结果：搬运自 http://poj.org/showmessage?message_id=141734
Case 1: 1
Case 2: 2
Case 3: 4
Case 4: 1
Case 5: 1
Case 6: -1
Case 7: 3
Case 8: -1
Case 9: 2
Case 10: 1
Case 11: 1
Case 12: -1
Case 13: -1
Case 14: 2
Case 15: 1
Case 16: 1
Case 17: 1
Case 18: -1
Case 19: -1
Case 20: -1

代码不难，关键要把每个点的区域排序，也要注意不要忽视小区间的右端点。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 struct Node {
 4     double left,right;
 5 } node[1001];
 6 bool cmp(Node a,Node b) {
 7     return a.left<b.left;
 8 }
 9 int main() {
10     int n,d,x,y,num=0;
11     while(cin>>n>>d&&n!=0) {
12         num++;
13         int ans=1;
14         int flag=1;
15         for(int i=0; i<n; i++) {
16             cin>>x>>y;
17             if(y>d&&flag) {
18                 printf("Case %d: -1\n",num);
19                 flag=0;
20             }
21             if(flag) {
22                 node[i].left=x-sqrt(d*d-y*y);
23                 node[i].right=x+sqrt(d*d-y*y);
24             }
25         }
26         if(flag) {
27             sort(node,node+n,cmp);
28             double rr=node[0].right;
29             for(int i=0; i<n; i++) {
30                 rr=min(rr,node[i].right);
31                 if(node[i].left>rr) {
32                     ans++;
33                     rr=node[i].right;
34                 }
35             }
36             printf("Case %d: %d\n",num,ans);
37         }
38 
39     }
40 
41     return 0;
42 }

1328:Radar Installation（贪心算法）

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