POJ 1328 Radar Installation(贪心)
来源:互联网 发布:大股东知乎 编辑:程序博客网 时间:2024/06/06 05:48
Radar Installation
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 75586 Accepted: 16939
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
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
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
3 21 2-3 12 11 20 20 0
Sample Output
Case 1: 2Case 2: 1
题目大意:
有一排海岸线,海岸线平行于X轴,海岸线的附近有很多岛屿,对于海岸线给定岛屿的坐标。问题是需要安装一些雷达来监测这些岛屿,每个雷达的监测半径为m,然后需要在海岸线上安装雷达,问你最少需要安装多少个,可以把所有的岛屿都监测到,输出数量,如果做不到,输出-1
题目解析:
根据给定的岛屿坐标,我们可以计算出每个岛屿在海岸线上安装雷达的区间,意思就是在这个区间内的任何地方安装雷达都是可以检测到此岛屿。这样我们根据n个岛屿的坐标就得到了n个不同的区间在海岸线上,然后我们只贪心的去处理这些区间,让每个区间内至少都存在一个雷达就可以了,那么从第一个区间, 选择最右边一个点, 如果不选择第一个区间最右一个点(选择前面的点), 那么把它换成最右的点之后, 以前得到满足的区间, 现在仍然得到满足, 所以第一个区间的最右一个点为贪婪选择, 选择该点之后, 将得到满足的区间删掉,继续更新最右边的点,直到结束。
#include <algorithm>#include <iostream>#include <numeric>#include <cstring>#include <iomanip>#include <string>#include <vector>#include <cstdio>#include <queue>#include <stack>#include <cmath>#include <map>#include <set>#define LL long long#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1const LL MAX = 405;const double esp = 1e-6;const double PI = 3.1415926535898;const int INF = 0x3f3f3f3f;using namespace std;struct node{ double l,r;}arr[1005];bool cmp(node a,node b){ return a.l < b.l;}int main(){ int n,Cas = 1; double m,k,nowx,nowy; while(~scanf("%d %lf",&n,&m) && (n||m)){ int f = 0; for(int i=0;i<n;i++){ scanf("%lf %lf",&nowx,&nowy); if(nowy > m) f = 1; arr[i].l = nowx - sqrt(m * m - nowy * nowy); arr[i].r = nowx + sqrt(m * m - nowy * nowy); } sort(arr,arr+n,cmp); int ans = 1; k = arr[0].r; for(int i=0;i<n-1;i++){ if(arr[i+1].l > k){ k = arr[i+1].r; ans += 1; } else if(arr[i+1].r < k){ k = arr[i+1].r; } } if(f) printf("Case %d: -1\n",Cas++); else printf("Case %d: %d\n",Cas++,ans); } return 0;}
0 0
- Radar Installation(POJ 1328)(贪心)
- POJ 1328 - Radar Installation(贪心)
- POJ 1328 Radar Installation(贪心)
- POJ 1328 Radar Installation(贪心)
- poj 1328 Radar Installation (贪心)
- poj 1328 Radar Installation (贪心)
- poj 1328 Radar Installation(贪心)
- poj 1328 Radar Installation (贪心)
- POJ 题目1328 Radar Installation(贪心)
- POJ 1328 Radar Installation(贪心)
- poj 1328 Radar Installation (贪心)
- POJ 1328 Radar Installation(贪心)
- POJ 1328 Radar Installation(经典贪心)
- POJ 1328 Radar Installation (贪心)
- POJ 1328 Radar Installation(贪心)
- POJ 1328 Radar Installation(贪心)
- POJ 1328 Radar Installation(区间贪心)
- POJ-1328Radar Installation(贪心)
- RPG游戏弹出框方法的封装
- 说说.NET中被我忽视的方法
- 【杭电】[5747]Aaronson
- 第一个工程3-代码解释二 ,检查球的的静止和一杆的结束
- C++ 指针
- POJ 1328 Radar Installation(贪心)
- CentOS的KVM实践(虚拟机创建、网桥配置、Spice)
- NSArray类的分析
- 【杭电】[5748]Bellovin
- Intellij打包Maven项目中那些七七八八
- PullToRefresh下拉刷新使用详解
- C++中类的静态成员
- Spring改版后的下载
- notepad++左侧显示目录树,安装插件