POJ 1328 Radar Installation(贪心)

Radar Installation

Time Limit: 1000MS

 

Memory Limit: 10000K

Total Submissions: 64148

 

Accepted: 14422

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

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

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

 

 

 

已疯，常年POJ，不成神，必得神经病。

 

首先吐槽下hdu的Web-DIY的渣渣测试数据，和炒鸡不稳定的系统。一份代码居然出现了RE，超时，AC三个结果，特么在逗我吧！！！

 

POJ数据太特么变态了，已哭。。。。。

 

解题思路：
将二维坐标转换为一维坐标上的区间。求最小点覆盖所有区间。

如果小岛能被雷达覆盖，则转换在一维坐标上的最大区间为，[x-sqrt(d*d*1.0-y*y),x+sqrt(d*d*1.0-y*y)]
如果其中某个小岛不能被覆盖，则用标记变量标记。直接输出-1。

如果所有小岛都能被雷达覆盖，则进行贪心策略。
对所有区间右端进行从小大到排序（右端相同时，左端从大到小排序），则如果出现区间包含的情况，小区间一定排在前面。
然后开始遍历区间。如果当前区间超过上一个区间的覆盖范围，则雷达数+1。

第一个区间去最右值。
证明：如果第一个区间不取最右值，而取中间的点，那么把点移到最右端，被满足的区间增加了。而原先被满足的区间现在一定被满足。

 

 

#include <cstdio>#include <cmath>#include <algorithm>const int maxn = 1010;using namespace std;struct radar{double s,e;}arr[maxn];bool cmp(radar x,radar y){if(x.e==y.e) return x.s>y.s;else return x.e<y.e;}int main(){int n,d,i,x,y,flag,count;int k=1;double temp;while(scanf("%d%d",&n,&d)){if(n==0&&d==0) break;flag=0;//标记是否有小岛无法被覆盖for(i=0;i<n;++i){scanf("%d%d",&x,&y);if(y>d){flag=1;//如果雷达不足以覆盖，则标记 continue; }temp=sqrt(d*d*1.0-y*y);arr[i].s=x-temp;arr[i].e=x+temp;}printf("Case %d: ",k++);if(flag){printf("-1\n");continue;}sort(arr,arr+n,cmp);count = 1;temp=arr[0].e;for(i=1;i<n;++i){if(arr[i].s>temp){temp=arr[i].e;count++;}}printf("%d\n",count);}return 0;}