NYOJ-287 Radar（贪心）

最新推荐文章于 2018-04-21 14:24:57 发布

原创最新推荐文章于 2018-04-21 14:24:57 发布 · 871 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#贪心

贪心专栏收录该内容

4 篇文章

订阅专栏

本文介绍了一种使用贪心算法解决雷达安装最小数量的问题。通过将岛屿视为坐标系中的点，并利用雷达覆盖范围，转换为x轴上的区间覆盖问题。具体步骤包括计算岛屿与海岸线形成的区间的左右端点、按区间右端点排序并采用贪心策略确定最少雷达数量。

Radar

时间限制： 1000 ms | 内存限制： 65535 KB

难度： 3

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.

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

来源

Beijing 2002

上传者

ctest

思路：

贪心算法；

1. 把小岛看做圆心, 雷达的半径作为半径, 求这个圆与x轴的交点x, y; 那在x,y之间任意位置放置雷达都能覆盖小岛.既把小岛的坐标(xi,yi)转化成在x轴上的闭区间(x,y);

2. 然后对所有小岛根据x升序排序,因为下一步放置雷达是从最左边往右放的;

3. 令雷达的左端点x = 第一个小岛的y,然后不断向后覆盖其后小岛的左端点x,此时可能出现三种情况：

a。雷达右端点y>小岛的左端点x,即此雷达可以覆盖该小岛，则继续；

b。如果雷达的左端点x < 小岛的x坐标，雷达肯定覆盖能此小岛,直接跳过。

c。如果雷达的右端点y < 小岛的左端点x，雷达的计数器加1，开始新的一轮循环；

代码:

#include<stdio.h>
#include<math.h>
#include<algorithm>

using namespace std;
const int MAXN=1005;

struct Line
{
	double l,r;
}line[MAXN];//每个岛作半径为d的圆，截得区间

bool cmp(Line a, Line b)
{
	return a.r < b.r;
}

int main()
{
	int n, d;
	int i, x, y;
	int num=1, flag, count;
	while(scanf("%d %d",&n, &d) && n && d){
		flag = 1;
		for(i = 0; i < n; i ++){
			scanf("%d %d",&x, &y);
			if(!flag) continue;
			if(y <= d){
				line[i].l = (double)x - sqrt((double)d * d - y * y);
				line[i].r = (double)x + sqrt((double)d * d - y * y);
			}
			else flag = 0;
		}
		if(!flag){
			printf("Case %d: -1\n",num++);
			continue;
		}
		sort(line, line+n, cmp);
		count = 1;
		double now = line[0].r;
		for(i = 1; i < n; i ++){
			if(line[i].l <= now + 0.00005) continue;
			now = line[i].r;
			count ++;
		}
		printf("Case %d: %d\n", num++, count);
	}

	return 0;
}