POJ 1328 Radar Installation 雷达安装贪心问题求解

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题目链接： POJ 1328 Radar Installation

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

Sample Output

Case 1: 2
Case 2: 1

Source

Beijing 2002

题意

在x轴表示的海岸线上需要布置雷达，以覆盖海中的n个海岛，雷达的覆盖半径为d，求解如何选择布置点才能使用到的雷达最少。如果有的海岛不能被覆盖到，那么输出-1。

分析

考察以海岛为圆心，做半径为d的圆，看与x轴相交的那段区间，这样的话，这个区间内的任何位置布置雷达，都是可以覆盖这个海岛的，对于所有的海岛，当然不乏求到的区间有部分重合的情况，那么在这个重合的区间中布置雷达，当然就能覆盖到两个以上的点，这样就能节省雷达，雷达所在的区间越多，节省的雷达就越多。

思想

典型的贪心思想。对于求出的每一个区间，我们进行排序，让区间右端点小的排在前面，如果右端点相等，那么左端点大的排在前面（想想为什么）。

那么该如何选择布置点呢？

首先我们选取排序号后的第一个区间的右端点为第一个布置点，它的位置为st，然后再按顺序找后面的区间。如果当前区间的左端点大于st，说明st位置的雷达不能覆盖到这个海岛，那么雷达数加1，同时更新st为这个区间的右端点。如果当前区间的左端点小于等于st，则说明st位置的雷达能覆盖这个海岛，忽略这个区间。

代码

/*POJ_1328_Radar_InstallationAuthor: Sign_Greedy
*/
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;int n, d;
struct seg
{double l, r;
}SEG[1010];seg pos(int x, int y)
{seg s;s.l = x - sqrt(d*d - y*y);s.r = x + sqrt(d*d - y*y);return s;
}
bool cmp(seg a, seg b)
{if(a.r == b.r) return a.l > b.l;return a.r < b.r;
}
int main()
{int x[1010], y[1010], cas = 1;while(scanf("%d%d", &n, &d), n, d){bool flag = true;for(int i = 0; i < n; i++){scanf("%d%d", &x[i], &y[i]);if(y[i] > d)flag = false;}if(!flag){printf("Case %d: -1\n", cas++);continue;}for(int i = 0; i < n; i++)SEG[i] = pos(x[i], y[i]);sort(SEG, SEG+n, cmp);int cnt = 1;double st = SEG[0].r;for(int i = 1; i < n; i++)if(SEG[i].l > st){st = SEG[i].r;cnt++;}printf("Case %d: %d\n", cas++, cnt);}return 0;
}

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