【POJ】1328 - Radar Installation（贪心）

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Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 75812 Accepted: 16983

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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1

Source

Beijing 2002

首先对于每一个岛，可以用 x ± sqrt （d * d - y * y ）求出的这个区间表示可以覆盖这个岛的雷达的位置。当然如果 y > d 的话这个岛就没法覆盖，就直接输出 -1 结束了。

然后对于这么些区间，如果有某些区间是有重叠的，那么这些岛放置一个雷达就可以全部覆盖，现在问题就是怎么解决区间覆盖了。

我们把雷达的结束位置排一下序（sort），然后从头开始，雷达肯定是放在结束的位置才最有利于后面的岛来 “ 蹭 ” 雷达。然后用一个数组记录一下哪些岛已经被雷达信号覆盖，对于已经被覆盖的，可以直接跳过。

代码如下：

#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>using namespace std;struct node{double st,endd;}radar[1011];bool cmp(node a , node b){return a.endd < b.endd;}int main(){int n,d;int Case = 1;bool ans;bool used[1011];int ant;while (~scanf ("%d %d",&n,&d) && (n || d)){ans = true;for (int i = 0 ; i < n ; i++){double x,y;scanf ("%lf %lf",&x,&y);if (y > d){ans = false;continue;}double t = sqrt(d*d - y*y);radar[i].st = x - t;radar[i].endd = x + t;}printf ("Case %d: ",Case++);if (!ans){printf ("-1\n");continue;}ant = 0;sort(radar , radar + n , cmp);//在结束点安装，所以按结束位置排序 memset (used , false , sizeof(used));for (int i = 0 ; i < n ; i++){if (used[i])continue;ant++;for (int j = i + 1 ; j < n ; j++){if (used[j])continue;if (radar[j].st <= radar[i].endd)used[j] = true;}}printf ("%d\n",ant);}return 0;}