贪心 Radar Installation (求最少探测雷达)
Description
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 is terminated by a line containing pair of zeros
Output
Sample Input
3 2
1 2
-3 1
2 1 1 2
0 2 0 0
Sample Output
Case 1: 2
Case 2: 1 先算出每个岛放探测雷达的x坐标范围,b为雷达探测半径,岛的坐标为(x,y),雷达的坐标范围是(x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)),定义sum=1,i从1开始,每个坐标范围左右边界与上一个坐标范围的右边界比较,具体看代码。
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
struct stu
{
double l,r;
} st[];
bool cmp(stu a,stu b)
{
return a.l<b.l;
}
int main()
{
int a,s,k=,sum,i;
double b,x,y,sky;
while(scanf("%d %lf",&a,&b) &&!(a==&&b==))
{
s=;
k++;
for(i = ; i < a ; i++)
{
scanf("%lf %lf",&x,&y);
if(b < || y > b)
{
s=-;
}
st[i].l=x-sqrt(b*b-y*y);
st[i].r=x+sqrt(b*b-y*y);
}
if(s == -) printf("Case %d: -1\n",k);
else
{
sort(st,st+a,cmp);
sum=;
sky=st[].r;
for(i = ; i < a ; i++)
{
if(st[i].r <= sky)
{
sky=st[i].r;
}
else
{
if(st[i].l > sky)
{
sky=st[i].r;
sum++;
}
}
}
printf("Case %d: %d\n",k,sum);
} }
}
贪心 Radar Installation (求最少探测雷达)的更多相关文章
- [ACM_贪心] Radar Installation
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28415#problem/A 题目大意:X轴为海岸线可放雷达监测目标点,告诉n个目标点和雷 ...
- 贪心 POJ 1328 Radar Installation
题目地址:http://poj.org/problem?id=1328 /* 贪心 (转载)题意:有一条海岸线,在海岸线上方是大海,海中有一些岛屿, 这些岛的位置已知,海岸线上有雷达,雷达的覆盖半径知 ...
- POJ 1328 Radar Installation 【贪心 区间选点】
解题思路:给出n个岛屿,n个岛屿的坐标分别为(a1,b1),(a2,b2)-----(an,bn),雷达的覆盖半径为r 求所有的岛屿都被覆盖所需要的最少的雷达数目. 首先将岛屿坐标进行处理,因为雷达的 ...
- poj 1328 Radar Installation (简单的贪心)
Radar Installation Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 42925 Accepted: 94 ...
- Radar Installation POJ - 1328(贪心)
Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. ...
- poj 1328 Radar Installation(贪心+快排)
Description Assume the coasting is an infinite straight line. Land is in one side of coasting, sea i ...
- POJ 1328 Radar Installation(很新颖的贪心,区间贪心)
Radar Installation Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 106491 Accepted: 2 ...
- 【OpenJ_Bailian - 1328】Radar Installation (贪心)
Radar Installation 原文是English,直接上中文 Descriptions: 假定海岸线是无限长的直线.陆地位于海岸线的一侧,海洋位于另一侧.每个小岛是位于海洋中的一个点.对于任 ...
- poj 1328 Radar Installation(nyoj 287 Radar):贪心
点击打开链接 Radar Installation Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 43490 Accep ...
随机推荐
- LuoguP1063 能量项链【区间Dp】
前天测试考了QAQ.这么经典的区间dp不再开一文好像有点可惜. 内容是从模拟赛结题报告中copy来的.QAQ noip能量项链原题. https://www.luogu.org/problemnew/ ...
- A - Supercentral Point CodeForces - 165A
One day Vasya painted a Cartesian coordinate system on a piece of paper and marked some set of point ...
- NOI2018D2T1 屠龙勇士
安利一下松松松的OJ: 传送门 Description: 有N条巨龙, 对于每个龙含有\(a_i\)的生命, 你有N + M把砍刀, 其中M把是直接给你的, N把是杀死对应的巨龙才能获得的, 每把 ...
- hdu 4565 So Easy! (共轭构造+矩阵快速幂)
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4565 题目大意: 给出a,b,n,m,求出的值, 解题思路: 因为题目中出现了开根号,和向上取整后求 ...
- how-to-fix-fs-re-evaluating-native-module-sources-is-not-supported-graceful
http://stackoverflow.com/questions/37346512/how-to-fix-fs-re-evaluating-native-module-sources-is-not ...
- C#---数据库访问通用类、Access数据库操作类、mysql类 .[转]
原文链接 //C# 数据库访问通用类 (ADO.NET)using System;using System.Collections.Generic;using System.Text;using Sy ...
- 移动端rem单位用法
1.rem(font size of the root element)是指相对于根元素的字体大小的单位,em(font size of the element)是指相对于父元素的字体大小的单位.它们 ...
- AJPFX理解反射及反射的应用
怎么理解反射,反射的应用 反射就是把Java类中的各种成分映射成相应的Java类. 一般情况下我们要解决某个问题,先找到相关的类,创建该类的对象,然后通过该对象调用对应的方 ...
- eclipse debug java 源码
当我们需要研究java SE的时候,debug 源码是个不错的选择,可以帮助我们清楚了解java 封装jar包的具体实现. 因为oracle 提供的源码jar包为了节省空间,所以没有将调试信息一起打包 ...
- iOS Programming Dynamic Type 2
iOS Programming Dynamic Type 2 You will need to update two parts of this view controller for ...