Radar(nyoj287)贪心

最新推荐文章于 2018-07-12 16:17:00 发布

橘子唯一

最新推荐文章于 2018-07-12 16:17:00 发布

阅读量550

点赞数

CC 4.0 BY-SA版权

分类专栏：贪心算法

本文链接：https://blog.youkuaiyun.com/u012584987/article/details/21477001

贪心算法专栏收录该内容

2 篇文章

订阅专栏

本文介绍了一个关于雷达安装数量最小化的算法问题。问题设定为在无限长的海岸线上，需要安装雷达来覆盖海上的岛屿，每个雷达都有固定的覆盖范围。目标是最小化所需雷达的数量以覆盖所有岛屿。

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

题目大意：输入小岛的个数n，雷达的半径r，然后输入小岛的位置坐标。雷达只能建在海岸线（即x轴）上。问最少需要几个雷达才能覆盖所有小岛。

#include <stdio.h>
#include <stdlib.h>
#include <memory.h>
#include <math.h>
#define N 1000

typedef struct 
{
	double left,right;
}Point;

int cmp(const void *a,const void *b)
{
	return (*(Point *)a).left - (*(Point *)b).left;
}
int main()
{
	int n,i,count,num=1;
	double x[N],y[N],min,r;
	double len;
	Point point[N];
	while(scanf("%d%lf",&n,&r))
	{
		if(n == 0 && r == 0)
			break;
		memset(point,0,sizeof(point));
		for(i=0; i<n; i++)
		{
			scanf("%lf%lf",&x[i],&y[i]);
			if(y[i] > r)
			{
				printf("Case %d:-1\n",num);
				num ++;
				break;
			}
			len = sqrt(r*r-y[i] *y[i]);
			point[i].right = x[i] + len;
			point[i].left = x[i] - len;
		}
		qsort(point,n,sizeof(Point),cmp);

		count = 1;//初始雷达数为1，覆盖第一个小岛
		min = point[0].right;//以第一个小岛与x轴的右交点为标准开始
		for(i=1; i<n; i++)
		{
			if(point[i].left > min)//如果后一个小岛的左交点大于前一个的右交点，说明不相交
			{
				count ++;//需要多加一个雷达
				min = point[i].right;//再以下一个小岛的右交点为标准探测
			}
			else
			{
				if(point[i].right < min)//如果，前一个的右交点小于后一个的左交点，前一个的雷达可以覆盖后一个
				{
					//不需要增加一个雷达
					min = point[i].right;
				}
			}
		}
		printf("Case %d: %d\n",num,count);
		num ++;
	}
	return 0;
}