nyoj 287 Radar 【区间选点】

最新推荐文章于 2018-04-21 14:24:57 发布

笑着走完自己的路

最新推荐文章于 2018-04-21 14:24:57 发布

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分类专栏：贪心

本文链接：https://blog.youkuaiyun.com/chenzhenyu123456/article/details/42585591

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解决海岸线上雷达安装问题，确保所有岛屿被覆盖。通过计算每个岛屿的覆盖区间，使用贪心算法找到最少雷达安装数量。

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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

我至今不知道它的输出数据间为什么要空一行······一开始加了就错了，后来去了就过了。思路很简单：找出可以覆盖点的区间范围[start，end]，接下来就是区间找点问题。（建议做这道题前做一下nyoj贪心专题里面的891找点和1036非洲小孩）

#include<stdio.h>
#include<math.h>
#include<algorithm>
using namespace std;
struct island
{
    double start;
    double end;
}num[1010];
bool cmp(island a,island b)
{
    return (a.end<b.end);
}
int main()
{
    int m,i;
    double r,x,y,max;
    int exist;
    int sum;
    int j=1;
    while(scanf("%d %lf",&m,&r)&&(m!=0||r!=0))
    {
        exist=1;
        for(i=0;i<m;i++)
        {
            scanf("%lf %lf",&x,&y);
            num[i].start=x-sqrt(r*r-y*y);
            num[i].end=x+sqrt(r*r-y*y);
            if(y>r)
            {
                exist=0;
                break;
            }
        }
        if(exist==0)
        printf("-1\n");
        else
        {
            sort(num,num+m,cmp);
            max=num[0].end;
            sum=1;
            for(i=1;i<m;i++)
            {
                if(num[i].start>max)
                {
                    sum++;
                    max=num[i].end;
                }
            }
            printf("Case %d: %d\n",j++,sum);
        }
    }
    return 0;
}